【Leetcode】【Easy】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.

 

递归的解法：

用树递归的思想，将两个树以结点作为比较单位，只关注对当前位置结点的操作（是否都存在此结点，存在时值是否相同）。 

注意：

1、只关注单个结点，结点的孩子交给递归去做，这样可以最简化逻辑；

2、结点是否存在、结点值是否相等、结点是否有孩子，是三个不同的概念（“结点是否有孩子”的概念本题中未体现）；

 

 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 
11 class Solution {
12 public:
13     bool isSameTree(TreeNode *p, TreeNode *q) {
14         if (!p && !q)
15             return true;
16         
17         if ((p && !q) || (!p && q) || (p->val != q->val))
18             return false;
19             
20         return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
21     }
22 };

 

迭代的解法：

引用先序/中序/后序/按层等遍历二叉树的思想，用堆栈（数组）存放遍历到的结点，进栈出栈的同时进行比较。

中序遍历（其他方法类似）：

 

 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         vector<TreeNode *> stackP;
14         vector<TreeNode *> stackQ;
15         TreeNode *currentP = p;
16         TreeNode *currentQ = q;
17         
18         while (currentP || stackP.size()) {
19             if (!currentP && !currentQ) {
20                 currentP = stackP.back();
21                 currentQ = stackQ.back();
22                 stackP.pop_back();
23                 stackQ.pop_back();
24                 currentP = currentP->right; 
25                 currentQ = currentQ->right;
26                 continue;
27             }
28             
29             if ((!currentP && currentQ) || (currentP && !currentQ) ||\
30                 (currentP->val != currentQ->val))
31                 return false;
32             
33             if (currentP->val == currentQ->val) {
34                 stackP.push_back(currentP);
35                 stackQ.push_back(currentQ);
36                 currentP = currentP->left;
37                 currentQ = currentQ->left;
38             }
39         }
40         
41         if (!currentQ) {
42             return true;
43         } else {
44             return false;
45         }
46         
47     }
48 };

 

附录：

C++中vector、stack、queue用法和区别

递归/非递归遍历二叉树

posted @ 2014-11-18 23:51  胡潇  阅读(283)  评论(0)  编辑  收藏  举报