【LeetCode】101. Symmetric Tree

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.

提示：

此题要求判断一个二叉树是否是左右对称的，且期望能够同时尝试一下迭代的方法和递归的方法。递归的方法相对来说简单一些，也更易读懂。迭代的方法需要利用栈来实现。

代码：

递归方法：

/**
 * 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* root) {
        if (!root) return true;
        return childSymmetric(root->left, root->right);
    }
    
    bool childSymmetric(TreeNode* l, TreeNode* r) {
        if (!l && !r) return true;
        if (!l || !r) return false;
        return l->val == r->val && childSymmetric(l->left, r->right) && childSymmetric(l->right, r->left);
    }
};

迭代方法：

/**
 * 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* root) {
        if (!root) return true;
        TreeNode *l, *r;
        stack<TreeNode*> s;
        if (root->left) {
            if (!root->right) return false;
            s.push(root->left);
            s.push(root->right);
        } else if (root->right) return false;
        
        while(!s.empty()) {
            l = s.top();s.pop();
            r = s.top();s.pop();
            if (l->val != r->val) return false;
            
            if (l->left) {
                if (!r->right) return false;
                s.push(l->left);
                s.push(r->right);
            } else if (r->right) return false;
            
            if (l->right) {
                if (!r->left) return false;
                s.push(l->right);
                s.push(r->left);
            } else if (r->left) return false;
        }
        return true;
    }
};