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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.
思路一：递归思想。时间复杂度O(n)，空间复杂度O(logN)
1 /** 2 * Definition for binary tree 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 == NULL) return true; 14 return isSymmetric(root->left, root->right); 15 } 16 bool isSymmetric(TreeNode *root1, TreeNode *root2) { 17 if (!root1 && !root2) return true; 18 if (!root1 || !root2) return false; 19 if (root1->val != root2->val) return false; 20 21 return isSymmetric(root1->left, root2->right) && isSymmetric(root1->right, root2->left); 22 } 23 };
思路二：迭代。层次遍历的思想。时间复杂度O(n),空间复杂度O(logN）
1 /** 2 * Definition for binary tree 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 == NULL) return true; 14 queue<TreeNode *> q; 15 q.push(root->left); 16 q.push(root->right); 17 18 while (!q.empty()) { 19 TreeNode *p1 = q.front(); 20 q.pop(); 21 TreeNode *p2 = q.front(); 22 q.pop(); 23 24 if (!p1 && !p2) continue; 25 if (!p1 || !p2) return false; 26 if (p1->val != p2->val) return false; 27 28 q.push(p1->left); 29 q.push(p2->right); 30 31 q.push(p1->right); 32 q.push(p2->left); 33 } 34 35 return true; 36 } 37 38 };
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原文地址：http://www.cnblogs.com/vincently/p/4232530.html
