LeetCode Algorithms #101 <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.

思路：

我是用递归做的，我把树不断拆成左右树，然后比较值是不是相等。

解：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isSymmetric(TreeNode* left, TreeNode* right)    {        if(left == nullptr && right == nullptr)            return true;        if((left == nullptr && right != nullptr) || (left != nullptr && right == nullptr))            return false;                    return (left->val == right->val && isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left));    }        bool isSymmetric(TreeNode* root) {        if(root == nullptr)            return true;                    return isSymmetric(root->left,root->right);    }};