//

// Solution101.hpp

// Algorithm

//

// Created by Pancf on 2019/12/29.

// Copyright © 2019 Pancf. All rights reserved.

//

#ifndef Solution101_hpp

#define Solution101_hpp

#include <stdio.h>

class Solution101 {

/**

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

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,null,3,null,3] is not:

1

/ \

2 2

\ \

3 3

Note:

Bonus points if you could solve it both recursively and iteratively.

================================================================================================

Accept details:

Runtime: 4 ms, faster than 84.36% of C++ online submissions for Symmetric Tree.

Memory Usage: 14.7 MB, less than 84.75% of C++ online submissions for Symmetric Tree.

思路：这题用递归比迭代简单多了

*/

public:

struct TreeNode {

int val;

TreeNode *left;

TreeNode *right;

TreeNode(int x) : val(x), left(NULL), right(NULL) {}

};

bool isSymmetric(TreeNode* root);

static void test() {

TreeNode root{1};

TreeNode node1{2};

TreeNode node2{2};

TreeNode node3{3};

TreeNode node4{4};

TreeNode node5{4};

TreeNode node6{3};

root.left = &node1;

root.right = &node2;

node1.left = &node3;

node1.right = &node4;

node2.left = &node5;

node2.right = &node6;

Solution101 s = Solution101();

printf("%d\n", s.isSymmetric(&root));

}

};

#endif /* Solution101_hpp */