这道题目我最初想的太多了，导致做了好久，之后从网上一看，一下醒悟过来，不难。
题目：
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.
解决方案：Runtime: 191 ms
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if(p==null&&q==null)
return true;
if(p!=null&&q==null || p==null&&q!=null){//here can be if(p == null || q == null){,but it will take more time
return false;
}
if(p.val == q.val){
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
return false;
}
}
总结：这道题目注意几个判断就好了，其他的就是单纯的递归。
原文：http://www.cnblogs.com/Pillar/p/4309031.html