[leetcode-101]Symmetric Tree(c++）

问题描述：

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

[code]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.

代码如下: 递归版/4ms

[code]/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool checkSym(struct TreeNode* leftNode,struct TreeNode* rightNode){
    if(!leftNode || !rightNode)
        return leftNode==rightNode;
    if(leftNode->val != rightNode->val)
        return false;
    return checkSym(leftNode->right,rightNode->left) && checkSym(leftNode->left,rightNode->right);

}
bool isSymmetric(struct TreeNode* root) {
    if(!root)
        return true;
    return checkSym(root->left,root->right);
}

代码如下：迭代版/4ms

(维护一个队列或者栈)

[code]class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        queue<TreeNode*> queue;
        if(!root)
            return true;
        queue.push(root->left);
        queue.push(root->right);
        TreeNode *front,*second;

        while(!queue.empty()){
            front = queue.front();queue.pop();
            second = queue.front();queue.pop();
            if(!front || !second){
                if(front!=second)
                    return false;
                continue;
            }

            if(front->val!=second->val)
                return false;
            queue.push(front->right);
            queue.push(second->left);
            queue.push(front->left);
            queue.push(second->right);
        }
        return true;
    }
};