[LeetCode]Symmetric Tree

题目描述:(链接)

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:

递归版:

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

迭代版:

/**
 * 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 == nullptr) return true;
        stack<TreeNode *> cache;
        cache.push(root->left);
        cache.push(root->right);
        while (!cache.empty()) {
            auto p = cache.top(); cache.pop();
            auto q = cache.top(); cache.pop();
            
            if (!p && !q) continue;
            if (!p || !q) return false;
            if (p->val != q->val) return false;
            
            cache.push(p->left); cache.push(q->right);
            cache.push(p->right); cache.push(q->left);
        }
        
        return true;
    }
};