LeetCode: Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
地址：https://oj.leetcode.com/problems/symmetric-tree/
算法：写了一个非递归的算法。用BFS遍历，每到一层结束判断该层节点是否对称，其中对于空的节点用0X80000000表示。代码：
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 isSymmetric(TreeNode *root) {
13 if(!root) return true;
14 queue<TreeNode *> que;
15 int num_push = 0;
16 int num_pop = 0;
17 int last = 1;
18 que.push(root);
19 ++num_push;
20 vector<int> temp;
21 while(!que.empty()){
22 TreeNode *node_pop = que.front();
23 if(node_pop){
24 temp.push_back(node_pop->val);
25 que.push(node_pop->left);
26 ++num_push;
27 que.push(node_pop->right);
28 ++num_push;
29 }
30 else
31 temp.push_back(0x80000000);
32 que.pop();
33 ++num_pop;
34 if(num_pop == last){
35 int i = 0;
36 while(i < temp.size()/2 && temp[i] == temp[temp.size()-i-1]) ++i;
37 if(i < temp.size()/2) return false;
38 last = num_push;
39 temp.clear();
40 }
41 }
42 return true;
43 }
44
45 };
