Symmetric Tree - LintCode

描述
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
Bonus points if you could solve it both recursively and iteratively.

樣例
For example, this binary tree {1,2,2,3,4,4,3} is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following {1,2,2,#,3,#,3} is not:

    1
   / \
  2   2
   \   \
   3    3

思路
遞歸求解，利用修改先序遍歷，對於一個二叉樹，先訪問右子樹再訪問左子樹，
另一個二叉樹就按先序遍歷，若得到節點的值均相同，說明是對稱的。

#ifndef C1360_H
#define C1360_H
#include<iostream>
using namespace std;
class TreeNode{
public:
    int val;
    TreeNode *left, *right;
    TreeNode(int val){
        this->val = val;
        this->left = this->right = NULL;
    }
};
class Solution {
public:
    /**
    * @param root: root of the given tree
    * @return: whether it is a mirror of itself
    */
    bool isSymmetric(TreeNode * root) {
        // Write your code here
        return helper(root, root);
    }
    //對於兩顆二叉樹,一個先訪問左子樹再訪問右子樹
    //另一個先訪問右子樹再訪問左子樹,訪問到的節點值均相同可以說明對稱
    bool helper(TreeNode *root1, TreeNode *root2)
    {
        if (!root1&&!root2)
            return true;
        if (!root1 || !root2)
            return false;
        if (root1->val != root2->val)
            return false;
        return helper(root1->left, root2->right) && helper(root1->right, root2->left);
    }
};
#endif