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LeetCode:Same Tree

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本文地址

 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, as
14         // 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, as
14         // the same Solution instance will be reused for each test case.
15         queue<TreeNode*> 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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原文地址：https://www.cnblogs.com/TenosDoIt/p/3440753.html