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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * 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) {
        vector<int> arr(0);
        queue<TreeNode*> q;
        vector<int>::iterator index_begin,index_end;
        TreeNode *leaf;
        int cnt=1,i;
        int leaves=0;
        
        TreeNode *p = root;
        if(p==NULL) return true;
        q.push(p);
        while(!q.empty()){
            arr.clear();
            for(i=0,leaves=0;i<cnt;i++){
                p=q.front();
                arr.push_back(p->val);
                q.pop();
                if(p->val==-999) continue;
                if(p->left!=NULL) {
                    q.push(p->left);
                    leaves++;
                }
                else {
                    leaf=(TreeNode*)malloc(sizeof(TreeNode));
                    leaf->val=-999;
                    q.push(leaf);
                    leaves++;
                }
               
                if(p->right!=NULL) {
                    q.push(p->right);
                    leaves++;
                }
                else {
                    leaf=(TreeNode*)malloc(sizeof(TreeNode));
                    leaf->val=-999;
                    q.push(leaf);
                    leaves++;
                }
            }
            cnt=leaves;
            if(cnt%2!=0) return false;
            index_begin=arr.begin();
            index_end=arr.end()-1;
            while(index_begin<index_end){
                if(*index_begin!=*index_end) return false;
                index_begin++;
                index_end--;
            }
            
        }
        return true;
    }
};

Symmetric Tree

原文：http://blog.csdn.net/uj_mosquito/article/details/41967569