Is it possible to compare two binary trees in less than O(n log n) time?












7














I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time. 



 public class TreeNode {

  int val;

  TreeNode left;

  TreeNode right;

  TreeNode(int x) { val = x; }

 }



 class Solution {

  public boolean isSameTree(TreeNode p, TreeNode q) {



    if  ( p == null && q==null)

        return true;



    if (p == null || q == null) 

        return false;



    if ( (p.val == q.val) && isSameTree(p.left, q.left) && 

      isSameTree(p.right, q.right))

        return true;

    else 

        return false;  

   }   

  }


My code takes O(n log n) time.



How to approach reducing the time required?






java algorithm time-complexity binary-tree







share|improve this question
























  • If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    2 hours ago
















7














I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time. 



 public class TreeNode {

  int val;

  TreeNode left;

  TreeNode right;

  TreeNode(int x) { val = x; }

 }



 class Solution {

  public boolean isSameTree(TreeNode p, TreeNode q) {



    if  ( p == null && q==null)

        return true;



    if (p == null || q == null) 

        return false;



    if ( (p.val == q.val) && isSameTree(p.left, q.left) && 

      isSameTree(p.right, q.right))

        return true;

    else 

        return false;  

   }   

  }


My code takes O(n log n) time.



How to approach reducing the time required?






java algorithm time-complexity binary-tree







share|improve this question
























  • If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    2 hours ago














7












7








7


1





I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time. 



 public class TreeNode {

  int val;

  TreeNode left;

  TreeNode right;

  TreeNode(int x) { val = x; }

 }



 class Solution {

  public boolean isSameTree(TreeNode p, TreeNode q) {



    if  ( p == null && q==null)

        return true;



    if (p == null || q == null) 

        return false;



    if ( (p.val == q.val) && isSameTree(p.left, q.left) && 

      isSameTree(p.right, q.right))

        return true;

    else 

        return false;  

   }   

  }


My code takes O(n log n) time.



How to approach reducing the time required?






java algorithm time-complexity binary-tree







share|improve this question















I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time. 



 public class TreeNode {

  int val;

  TreeNode left;

  TreeNode right;

  TreeNode(int x) { val = x; }

 }



 class Solution {

  public boolean isSameTree(TreeNode p, TreeNode q) {



    if  ( p == null && q==null)

        return true;



    if (p == null || q == null) 

        return false;



    if ( (p.val == q.val) && isSameTree(p.left, q.left) && 

      isSameTree(p.right, q.right))

        return true;

    else 

        return false;  

   }   

  }


My code takes O(n log n) time.



How to approach reducing the time required?






java algorithm time-complexity binary-tree




java algorithm time-complexity binary-tree






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









nullpointer

43.8k1093179




43.8k1093179










asked 5 hours ago









Louise

361




361












  • If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    2 hours ago


















  • If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    2 hours ago
















If you happen to have a size variable at the base of the structure, compare that first.
– Boann
2 hours ago




If you happen to have a size variable at the base of the structure, compare that first.
– Boann
2 hours ago












1 Answer
1






active

oldest

votes


















8














The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.






share|improve this answer



















  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    4 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8














The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.






share|improve this answer



















  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    4 hours ago
















8














The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.






share|improve this answer



















  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    4 hours ago














8












8








8






The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.






share|improve this answer














The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.



Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).



Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.







share|improve this answer














share|improve this answer



share|improve this answer








edited 3 hours ago









ruakh

124k13197251




124k13197251










answered 4 hours ago









nullpointer

43.8k1093179




43.8k1093179








  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    4 hours ago














  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    4 hours ago








2




2




In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago




In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago


















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