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.

算法1：递归解法，如果根节点相同，再递归的看左子树和右子树是否相同                                              

1 /**  2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     bool isSameTree(TreeNode *p, TreeNode *q) {13         // IMPORTANT: Please reset any member data you declared, as14         // the same Solution instance will be reused for each test case.15         if(p && q)16         {17             if(p->val == q->val && isSameTree(p->left, q->left) && 18                 isSameTree(p->right, q->right))19                 return true;20             else return false;21         }22         else if(p || q)23             return false;24         else return true;25     }26 };

算法2：通过层序遍历，分别用两个队列保存两棵树的层序节点。每次从两个队列中分别取出一个元素，如果他们的值相同，再继续遍历子节点。注意每次取出的两个元素的左子树要么同时非空，要么同时为空，右子树也一样

1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     bool isSameTree(TreeNode *p, TreeNode *q) {13         // IMPORTANT: Please reset any member data you declared, as14         // the same Solution instance will be reused for each test case.15         queue
     
       pQueue, qQueue;16         if(p)pQueue.push(p);17         if(q)qQueue.push(q);18         while(pQueue.empty() == false && qQueue.empty() == false)19         {20             TreeNode *pp = pQueue.front();21             TreeNode *qq = qQueue.front();22             pQueue.pop();  qQueue.pop();23             if(pp->val == qq->val)24             {25                 if(pp->left && qq->left)26                 {27                     pQueue.push(pp->left);28                     qQueue.push(qq->left);29                 }30                 else if(pp->left || qq->left)31                     return false;32                 if(pp->right && qq->right)33                 {34                     pQueue.push(pp->right);35                     qQueue.push(qq->right);36                 }37                 else if(pp->right || qq->right)38                     return false;39             }40             else return false;41         }42         if(pQueue.empty() && qQueue.empty())43             return true;44         else return false;45     }46 };
     

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