[Leetcode 6] 100 Same Tree

Problem:

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

Analysis:

Use a queue to do level order traversal is not proper in this problem since this method ignores the structure information of the tree. One possible way to deal with this problem is recursion. Starting from the root, 1. check the existance of nodes, if both are null, return true; One is null, the other is not, return false; 2. check the value equivalence, if not equal, return false, else recursively check their left and right children's equivalence.

Then time complexity in worst case is O(2*min{m, n}) where we have to check all the nodes of smaller tree and corresponding nodes in the larger tree. The space complexity is O(n+m), where n is the size of the first tree and m is the size of the second tree.

Code:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        // Start typing your Java solution below
        // DO NOT write main() function

         if (p==null && q==null) 
              return true;
          else if ((p==null && q!=null) || (p!=null && q==null))
              return false;
          else if (p.val != q.val) 
              return false;
          else
              return (isSameTree(p.left, q.left)) 
                      && (isSameTree(p.right, q.right));
    }
}

Attention:

Here is the first wrong implementation of this problem:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        // Start typing your Java solution below
        // DO NOT write main() function

        if (p==null && q==null) 
            return true;
        else if ((p==null && q!=null) || 
                    (q!=null && q==null)) 
            return false;
        else if (p.val != q.val) 
            return false;
        else
            return (isSameTree(p.left, q.left)) 
                    && (isSameTree(p.right, q.right));
    }
}

There is a typo in line 18, where the condition should be (p!=null && q==null), be careful while typing.

Another thing is be careful with the comparision with null, it means that the given reference points to nothing, not the regerenced value is null. To check references value is equivalent or not, use ref1.equals(ref2) rather than ref1==ref2.