LeetCode: Symmetric Tree

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：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        if(!root)   return true;
        queue<TreeNode *> que;
        int num_push = 0;
        int num_pop  = 0;
        int last     = 1;
        que.push(root);
        ++num_push;
        vector<int> temp;
        while(!que.empty()){
            TreeNode *node_pop = que.front();
            if(node_pop){
                temp.push_back(node_pop->val);
                que.push(node_pop->left);
                ++num_push;
                que.push(node_pop->right);
                ++num_push;
            }
            else
                temp.push_back(0x80000000);
            que.pop();
            ++num_pop;
            if(num_pop == last){
                int i = 0;
                while(i < temp.size()/2 && temp[i] == temp[temp.size()-i-1])    ++i;
                if(i < temp.size()/2)   return false;
                last = num_push;
                temp.clear();
            }
        }
        return true;
    }
    
};