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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:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isSymmetric(TreeNode* root) { 13 if (root == nullptr) { 14 return true; 15 } 16 17 return isSymmetric(root->left, root->right); 18 } 19 20 bool isSymmetric(TreeNode *left, TreeNode *right) { 21 if (!left && !right) return true; 22 if (!left || !right) return false; 23 24 return left->val == right->val && 25 isSymmetric(left->left, right->right) && 26 isSymmetric(left->right, right->left); 27 } 28 };
迭代版:
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isSymmetric(TreeNode* root) { 13 if (root == nullptr) return true; 14 stack<TreeNode *> cache; 15 cache.push(root->left); 16 cache.push(root->right); 17 while (!cache.empty()) { 18 auto p = cache.top(); cache.pop(); 19 auto q = cache.top(); cache.pop(); 20 21 if (!p && !q) continue; 22 if (!p || !q) return false; 23 if (p->val != q->val) return false; 24 25 cache.push(p->left); cache.push(q->right); 26 cache.push(p->right); cache.push(q->left); 27 } 28 29 return true; 30 } 31 };
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原文地址：http://www.cnblogs.com/skycore/p/5021367.html
