Binary Tree Node Location and Helper Dictionary









up vote
1
down vote

favorite












Thanks for your help



Background



I have a question about locating nodes in a tree.



I have a simple binary tree. Each node has a piece of data in it. Lets say it looks like this:



 a
/
b c


Where a=root, b=root.left, c=root.right



The tree is not created manually. Let's say I get a request to add new_data to node c.



I'm confused about how to know where c is without explicitly writing root.right.data=new_data.



My first thought is to create some type of helper dictionary with references to the node locations, like:



helper = 
'a'= root,
'b'= root.left,
'c'= root.right



So that when I get a request, I can go to the helper and say something to the effect of:



helper.get('c').data=new_data



Questions



Am I in the right ballpark here? Recursively searching an entire tree repeatedly seems a bit much - this helper could be updated occasionally when the tree changes its node structure.



I'm confused about how to actually return my location for each node when I do recursively crawl the tree. How can I create this helper?










share|improve this question



















  • 1




    You might have a look at this answer: stackoverflow.com/questions/2598437/…
    – Ivonet
    Nov 11 at 19:52










  • Why you need the tree anyway ? why not just put your data in a dictionary and that's all ?
    – Schidu Luca
    Nov 11 at 19:55











  • @SchiduLuca I've simplified the example dramatically. The tree provides critical links between parent and child objects which can act on each other in different ways. The tree provides a very nice structure for this. I don't believe a dictionary would be productive.
    – JW2
    Nov 11 at 20:11










  • @JW2 Recursively searching an entire tree repeatedly seems a bit much . If you decided to put your data in such a form then it's ok to traverse recursively. If you think it's too much just don't use the tree at all..
    – Schidu Luca
    Nov 11 at 20:20















up vote
1
down vote

favorite












Thanks for your help



Background



I have a question about locating nodes in a tree.



I have a simple binary tree. Each node has a piece of data in it. Lets say it looks like this:



 a
/
b c


Where a=root, b=root.left, c=root.right



The tree is not created manually. Let's say I get a request to add new_data to node c.



I'm confused about how to know where c is without explicitly writing root.right.data=new_data.



My first thought is to create some type of helper dictionary with references to the node locations, like:



helper = 
'a'= root,
'b'= root.left,
'c'= root.right



So that when I get a request, I can go to the helper and say something to the effect of:



helper.get('c').data=new_data



Questions



Am I in the right ballpark here? Recursively searching an entire tree repeatedly seems a bit much - this helper could be updated occasionally when the tree changes its node structure.



I'm confused about how to actually return my location for each node when I do recursively crawl the tree. How can I create this helper?










share|improve this question



















  • 1




    You might have a look at this answer: stackoverflow.com/questions/2598437/…
    – Ivonet
    Nov 11 at 19:52










  • Why you need the tree anyway ? why not just put your data in a dictionary and that's all ?
    – Schidu Luca
    Nov 11 at 19:55











  • @SchiduLuca I've simplified the example dramatically. The tree provides critical links between parent and child objects which can act on each other in different ways. The tree provides a very nice structure for this. I don't believe a dictionary would be productive.
    – JW2
    Nov 11 at 20:11










  • @JW2 Recursively searching an entire tree repeatedly seems a bit much . If you decided to put your data in such a form then it's ok to traverse recursively. If you think it's too much just don't use the tree at all..
    – Schidu Luca
    Nov 11 at 20:20













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Thanks for your help



Background



I have a question about locating nodes in a tree.



I have a simple binary tree. Each node has a piece of data in it. Lets say it looks like this:



 a
/
b c


Where a=root, b=root.left, c=root.right



The tree is not created manually. Let's say I get a request to add new_data to node c.



I'm confused about how to know where c is without explicitly writing root.right.data=new_data.



My first thought is to create some type of helper dictionary with references to the node locations, like:



helper = 
'a'= root,
'b'= root.left,
'c'= root.right



So that when I get a request, I can go to the helper and say something to the effect of:



helper.get('c').data=new_data



Questions



Am I in the right ballpark here? Recursively searching an entire tree repeatedly seems a bit much - this helper could be updated occasionally when the tree changes its node structure.



I'm confused about how to actually return my location for each node when I do recursively crawl the tree. How can I create this helper?










share|improve this question















Thanks for your help



Background



I have a question about locating nodes in a tree.



I have a simple binary tree. Each node has a piece of data in it. Lets say it looks like this:



 a
/
b c


Where a=root, b=root.left, c=root.right



The tree is not created manually. Let's say I get a request to add new_data to node c.



I'm confused about how to know where c is without explicitly writing root.right.data=new_data.



My first thought is to create some type of helper dictionary with references to the node locations, like:



helper = 
'a'= root,
'b'= root.left,
'c'= root.right



So that when I get a request, I can go to the helper and say something to the effect of:



helper.get('c').data=new_data



Questions



Am I in the right ballpark here? Recursively searching an entire tree repeatedly seems a bit much - this helper could be updated occasionally when the tree changes its node structure.



I'm confused about how to actually return my location for each node when I do recursively crawl the tree. How can I create this helper?







python recursion binary-tree binary-search-tree






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share|improve this question













share|improve this question




share|improve this question








edited Nov 11 at 20:47









martineau

65.3k987176




65.3k987176










asked Nov 11 at 19:40









JW2

336




336







  • 1




    You might have a look at this answer: stackoverflow.com/questions/2598437/…
    – Ivonet
    Nov 11 at 19:52










  • Why you need the tree anyway ? why not just put your data in a dictionary and that's all ?
    – Schidu Luca
    Nov 11 at 19:55











  • @SchiduLuca I've simplified the example dramatically. The tree provides critical links between parent and child objects which can act on each other in different ways. The tree provides a very nice structure for this. I don't believe a dictionary would be productive.
    – JW2
    Nov 11 at 20:11










  • @JW2 Recursively searching an entire tree repeatedly seems a bit much . If you decided to put your data in such a form then it's ok to traverse recursively. If you think it's too much just don't use the tree at all..
    – Schidu Luca
    Nov 11 at 20:20













  • 1




    You might have a look at this answer: stackoverflow.com/questions/2598437/…
    – Ivonet
    Nov 11 at 19:52










  • Why you need the tree anyway ? why not just put your data in a dictionary and that's all ?
    – Schidu Luca
    Nov 11 at 19:55











  • @SchiduLuca I've simplified the example dramatically. The tree provides critical links between parent and child objects which can act on each other in different ways. The tree provides a very nice structure for this. I don't believe a dictionary would be productive.
    – JW2
    Nov 11 at 20:11










  • @JW2 Recursively searching an entire tree repeatedly seems a bit much . If you decided to put your data in such a form then it's ok to traverse recursively. If you think it's too much just don't use the tree at all..
    – Schidu Luca
    Nov 11 at 20:20








1




1




You might have a look at this answer: stackoverflow.com/questions/2598437/…
– Ivonet
Nov 11 at 19:52




You might have a look at this answer: stackoverflow.com/questions/2598437/…
– Ivonet
Nov 11 at 19:52












Why you need the tree anyway ? why not just put your data in a dictionary and that's all ?
– Schidu Luca
Nov 11 at 19:55





Why you need the tree anyway ? why not just put your data in a dictionary and that's all ?
– Schidu Luca
Nov 11 at 19:55













@SchiduLuca I've simplified the example dramatically. The tree provides critical links between parent and child objects which can act on each other in different ways. The tree provides a very nice structure for this. I don't believe a dictionary would be productive.
– JW2
Nov 11 at 20:11




@SchiduLuca I've simplified the example dramatically. The tree provides critical links between parent and child objects which can act on each other in different ways. The tree provides a very nice structure for this. I don't believe a dictionary would be productive.
– JW2
Nov 11 at 20:11












@JW2 Recursively searching an entire tree repeatedly seems a bit much . If you decided to put your data in such a form then it's ok to traverse recursively. If you think it's too much just don't use the tree at all..
– Schidu Luca
Nov 11 at 20:20





@JW2 Recursively searching an entire tree repeatedly seems a bit much . If you decided to put your data in such a form then it's ok to traverse recursively. If you think it's too much just don't use the tree at all..
– Schidu Luca
Nov 11 at 20:20













2 Answers
2






active

oldest

votes

















up vote
0
down vote













If you have a reference to node c, you can use it directly, like so c.data=new_data.



Otherwise, if what you get is e.g. a string, then:



  1. Is your tree sorted, or can it be sorted? (i.e. is it a BST?) If so, use it.

  2. If it's not, then is there any other property of the tree/nodes that you can use to prune the search? (i.e. you have some information to limit where to search). If so, use it. If you don't know if you can use it, you should specify whatever properties your data has in your question.

  3. If the node can be anywhere, then you pretty much do have to search the whole tree (since the node you want can be anywhere). If that's the case though, is there a purpose to the tree structure? Maybe a hash table might be useful instead. (You could index the tree after you get it)

btw. Note that searching the tree should be O(n) in size of the tree, up to you if that's excessive or not.






share|improve this answer




















  • it is a BST. Each node does have a unique identifier. I think you're right - whether I'm searching a helper dictionary for that key helper.get(key) or I'm traversing the tree I am still doing a search. It's not immediately evident to me how though to return the instance of a node when I search for it.
    – JW2
    Nov 11 at 20:14











  • Ah, in that case, you can refer to _find() method in this answer stackoverflow.com/a/28864021/272515 that Ivonet linked (note that it doesn't actually use self except to reference the _find method recursively. So it should be trivial to convert to a function for your purpose)
    – Eric
    Nov 11 at 21:03











  • You're right. Thanks for the help! See below
    – JW2
    Nov 11 at 21:54

















up vote
0
down vote













Answer



The answer is to recursively search. My thoughts of helper dict stem from my unfamiliarity w/ BST. I added the following to my nodes to search the tree preorder:



def find_node(self, start, id):
if start:
if start.id == id:
return start
else:
node = self.find_node(start.left, id)
if node:
return node
else:
node = self.find_node(start.right, id)
if node:
return node
else:
return None


This video was super helpful in helping me understand BST. I realized instead of just printing the traversal I could return the actual instance of the class itself when I found the right id.






share|improve this answer




















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    If you have a reference to node c, you can use it directly, like so c.data=new_data.



    Otherwise, if what you get is e.g. a string, then:



    1. Is your tree sorted, or can it be sorted? (i.e. is it a BST?) If so, use it.

    2. If it's not, then is there any other property of the tree/nodes that you can use to prune the search? (i.e. you have some information to limit where to search). If so, use it. If you don't know if you can use it, you should specify whatever properties your data has in your question.

    3. If the node can be anywhere, then you pretty much do have to search the whole tree (since the node you want can be anywhere). If that's the case though, is there a purpose to the tree structure? Maybe a hash table might be useful instead. (You could index the tree after you get it)

    btw. Note that searching the tree should be O(n) in size of the tree, up to you if that's excessive or not.






    share|improve this answer




















    • it is a BST. Each node does have a unique identifier. I think you're right - whether I'm searching a helper dictionary for that key helper.get(key) or I'm traversing the tree I am still doing a search. It's not immediately evident to me how though to return the instance of a node when I search for it.
      – JW2
      Nov 11 at 20:14











    • Ah, in that case, you can refer to _find() method in this answer stackoverflow.com/a/28864021/272515 that Ivonet linked (note that it doesn't actually use self except to reference the _find method recursively. So it should be trivial to convert to a function for your purpose)
      – Eric
      Nov 11 at 21:03











    • You're right. Thanks for the help! See below
      – JW2
      Nov 11 at 21:54














    up vote
    0
    down vote













    If you have a reference to node c, you can use it directly, like so c.data=new_data.



    Otherwise, if what you get is e.g. a string, then:



    1. Is your tree sorted, or can it be sorted? (i.e. is it a BST?) If so, use it.

    2. If it's not, then is there any other property of the tree/nodes that you can use to prune the search? (i.e. you have some information to limit where to search). If so, use it. If you don't know if you can use it, you should specify whatever properties your data has in your question.

    3. If the node can be anywhere, then you pretty much do have to search the whole tree (since the node you want can be anywhere). If that's the case though, is there a purpose to the tree structure? Maybe a hash table might be useful instead. (You could index the tree after you get it)

    btw. Note that searching the tree should be O(n) in size of the tree, up to you if that's excessive or not.






    share|improve this answer




















    • it is a BST. Each node does have a unique identifier. I think you're right - whether I'm searching a helper dictionary for that key helper.get(key) or I'm traversing the tree I am still doing a search. It's not immediately evident to me how though to return the instance of a node when I search for it.
      – JW2
      Nov 11 at 20:14











    • Ah, in that case, you can refer to _find() method in this answer stackoverflow.com/a/28864021/272515 that Ivonet linked (note that it doesn't actually use self except to reference the _find method recursively. So it should be trivial to convert to a function for your purpose)
      – Eric
      Nov 11 at 21:03











    • You're right. Thanks for the help! See below
      – JW2
      Nov 11 at 21:54












    up vote
    0
    down vote










    up vote
    0
    down vote









    If you have a reference to node c, you can use it directly, like so c.data=new_data.



    Otherwise, if what you get is e.g. a string, then:



    1. Is your tree sorted, or can it be sorted? (i.e. is it a BST?) If so, use it.

    2. If it's not, then is there any other property of the tree/nodes that you can use to prune the search? (i.e. you have some information to limit where to search). If so, use it. If you don't know if you can use it, you should specify whatever properties your data has in your question.

    3. If the node can be anywhere, then you pretty much do have to search the whole tree (since the node you want can be anywhere). If that's the case though, is there a purpose to the tree structure? Maybe a hash table might be useful instead. (You could index the tree after you get it)

    btw. Note that searching the tree should be O(n) in size of the tree, up to you if that's excessive or not.






    share|improve this answer












    If you have a reference to node c, you can use it directly, like so c.data=new_data.



    Otherwise, if what you get is e.g. a string, then:



    1. Is your tree sorted, or can it be sorted? (i.e. is it a BST?) If so, use it.

    2. If it's not, then is there any other property of the tree/nodes that you can use to prune the search? (i.e. you have some information to limit where to search). If so, use it. If you don't know if you can use it, you should specify whatever properties your data has in your question.

    3. If the node can be anywhere, then you pretty much do have to search the whole tree (since the node you want can be anywhere). If that's the case though, is there a purpose to the tree structure? Maybe a hash table might be useful instead. (You could index the tree after you get it)

    btw. Note that searching the tree should be O(n) in size of the tree, up to you if that's excessive or not.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 11 at 19:56









    Eric

    1,06221124




    1,06221124











    • it is a BST. Each node does have a unique identifier. I think you're right - whether I'm searching a helper dictionary for that key helper.get(key) or I'm traversing the tree I am still doing a search. It's not immediately evident to me how though to return the instance of a node when I search for it.
      – JW2
      Nov 11 at 20:14











    • Ah, in that case, you can refer to _find() method in this answer stackoverflow.com/a/28864021/272515 that Ivonet linked (note that it doesn't actually use self except to reference the _find method recursively. So it should be trivial to convert to a function for your purpose)
      – Eric
      Nov 11 at 21:03











    • You're right. Thanks for the help! See below
      – JW2
      Nov 11 at 21:54
















    • it is a BST. Each node does have a unique identifier. I think you're right - whether I'm searching a helper dictionary for that key helper.get(key) or I'm traversing the tree I am still doing a search. It's not immediately evident to me how though to return the instance of a node when I search for it.
      – JW2
      Nov 11 at 20:14











    • Ah, in that case, you can refer to _find() method in this answer stackoverflow.com/a/28864021/272515 that Ivonet linked (note that it doesn't actually use self except to reference the _find method recursively. So it should be trivial to convert to a function for your purpose)
      – Eric
      Nov 11 at 21:03











    • You're right. Thanks for the help! See below
      – JW2
      Nov 11 at 21:54















    it is a BST. Each node does have a unique identifier. I think you're right - whether I'm searching a helper dictionary for that key helper.get(key) or I'm traversing the tree I am still doing a search. It's not immediately evident to me how though to return the instance of a node when I search for it.
    – JW2
    Nov 11 at 20:14





    it is a BST. Each node does have a unique identifier. I think you're right - whether I'm searching a helper dictionary for that key helper.get(key) or I'm traversing the tree I am still doing a search. It's not immediately evident to me how though to return the instance of a node when I search for it.
    – JW2
    Nov 11 at 20:14













    Ah, in that case, you can refer to _find() method in this answer stackoverflow.com/a/28864021/272515 that Ivonet linked (note that it doesn't actually use self except to reference the _find method recursively. So it should be trivial to convert to a function for your purpose)
    – Eric
    Nov 11 at 21:03





    Ah, in that case, you can refer to _find() method in this answer stackoverflow.com/a/28864021/272515 that Ivonet linked (note that it doesn't actually use self except to reference the _find method recursively. So it should be trivial to convert to a function for your purpose)
    – Eric
    Nov 11 at 21:03













    You're right. Thanks for the help! See below
    – JW2
    Nov 11 at 21:54




    You're right. Thanks for the help! See below
    – JW2
    Nov 11 at 21:54












    up vote
    0
    down vote













    Answer



    The answer is to recursively search. My thoughts of helper dict stem from my unfamiliarity w/ BST. I added the following to my nodes to search the tree preorder:



    def find_node(self, start, id):
    if start:
    if start.id == id:
    return start
    else:
    node = self.find_node(start.left, id)
    if node:
    return node
    else:
    node = self.find_node(start.right, id)
    if node:
    return node
    else:
    return None


    This video was super helpful in helping me understand BST. I realized instead of just printing the traversal I could return the actual instance of the class itself when I found the right id.






    share|improve this answer
























      up vote
      0
      down vote













      Answer



      The answer is to recursively search. My thoughts of helper dict stem from my unfamiliarity w/ BST. I added the following to my nodes to search the tree preorder:



      def find_node(self, start, id):
      if start:
      if start.id == id:
      return start
      else:
      node = self.find_node(start.left, id)
      if node:
      return node
      else:
      node = self.find_node(start.right, id)
      if node:
      return node
      else:
      return None


      This video was super helpful in helping me understand BST. I realized instead of just printing the traversal I could return the actual instance of the class itself when I found the right id.






      share|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Answer



        The answer is to recursively search. My thoughts of helper dict stem from my unfamiliarity w/ BST. I added the following to my nodes to search the tree preorder:



        def find_node(self, start, id):
        if start:
        if start.id == id:
        return start
        else:
        node = self.find_node(start.left, id)
        if node:
        return node
        else:
        node = self.find_node(start.right, id)
        if node:
        return node
        else:
        return None


        This video was super helpful in helping me understand BST. I realized instead of just printing the traversal I could return the actual instance of the class itself when I found the right id.






        share|improve this answer












        Answer



        The answer is to recursively search. My thoughts of helper dict stem from my unfamiliarity w/ BST. I added the following to my nodes to search the tree preorder:



        def find_node(self, start, id):
        if start:
        if start.id == id:
        return start
        else:
        node = self.find_node(start.left, id)
        if node:
        return node
        else:
        node = self.find_node(start.right, id)
        if node:
        return node
        else:
        return None


        This video was super helpful in helping me understand BST. I realized instead of just printing the traversal I could return the actual instance of the class itself when I found the right id.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 11 at 21:53









        JW2

        336




        336



























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