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100. Same Tree

时间:2016-06-26 22:38:49      阅读:276      评论:0      收藏:0      [点我收藏+]

1. 问题描述

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.

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Tags: Tree Depth-first Search
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

2. 解题思路


3. 代码

 1 class Solution {
 2 public:
 3     bool isSameTree(TreeNode* p, TreeNode* q)
 4     {
 5         if (NULL == p && NULL == q)
 6         {
 7             return true;
 8         }
 9 
10         if (NULL != p && NULL != q)
11         {
12             if (p->val == q->val)
13             {
14                 return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
15             }
16             else
17             {
18                 return false;
19             }
20         }
21         return false;
22     }
23 #pragma region 非递归方式
24     bool isSameTree_1(TreeNode *p, TreeNode *q) 
25     {
26         if(!isSameNode(p, q))
27             return false;
28         if(!p && !q)
29             return true;
30 
31         queue<TreeNode*> lqueue;
32         queue<TreeNode*> rqueue;
33         lqueue.push(p);
34         rqueue.push(q);
35         while(!lqueue.empty() && !rqueue.empty())
36         {
37             TreeNode* lfront = lqueue.front();
38             TreeNode* rfront = rqueue.front();
39 
40             lqueue.pop();
41             rqueue.pop();
42 
43             if(!isSameNode(lfront->left, rfront->left))
44                 return false;
45             if(lfront->left && rfront->left)
46             {
47                 lqueue.push(lfront->left);
48                 rqueue.push(rfront->left);
49             }
50 
51             if(!isSameNode(lfront->right, rfront->right))
52                 return false;
53             if(lfront->right && rfront->right)
54             {
55                 lqueue.push(lfront->right);
56                 rqueue.push(rfront->right);
57             }
58         }
59         return true;
60     }
61     bool isSameNode(TreeNode* p, TreeNode *q)
62     {
63         if(!p && !q)
64             return true;
65         if((p && !q) || (!p && q) || (p->val != q->val))
66             return false;
67         return true;
68     }
69 #pragma endregion
70 };

4. 反思

100. Same Tree

原文:http://www.cnblogs.com/whl2012/p/5596735.html

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