LeetCode OJ

题目：

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.

解题思路：

递归：左指针和右指针，对称递归，即“左的左”和“右的右”对应，“左的右”和“右的左”对应。

非递归：中序遍历左子树，将结果保存起来，中序遍历右子树将结果保存起来；顺序比较每个元素是否相等。

代码：

/**
 * 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 judge(TreeNode *p_left, TreeNode *p_right) {
        if (p_left == NULL && p_right == NULL) {
            return true;
        } else if ((p_left != NULL && p_right == NULL) || (p_left == NULL && p_right != NULL)) {
            return false; 
        }
        
        if (p_left->val != p_right->val) return false;
        
        if ((p_left->left != p_right->right && (p_left->left == NULL || p_right->right == NULL)) ||
            (p_left->right != p_right->left && (p_left->right == NULL || p_right->left == NULL))) {
                return false;
        }

        return judge(p_left->left, p_right->right) && judge(p_left->right, p_right->left); 
        
    }
    bool isSymmetric(TreeNode *root) {
        if (root == NULL || (root->left == NULL && root->right == NULL)) {
            return true;
        } else if (root->left == NULL || root->right == NULL) {
            return false;
        } 
        if (root->left->val == root->right->val) {
            return judge(root->left, root->right);
        }
        return false;
    }
};