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.
分类:Tree BFS DFS
代码:
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 13 bool isSymmetric(TreeNode *root) { 14 if (!root) return true; 15 return helper(root->left, root->right); 16 } 17 18 bool helper(TreeNode* p, TreeNode* q) { 19 if (!p && !q) { 20 return true; 21 } else if (!p || !q) { 22 return false; 23 } 24 25 if (p->val != q->val) { 26 return false; 27 } 28 29 return helper(p->left,q->right) && helper(p->right, q->left); 30 } 31 };
本文探讨了如何检查一个二叉树是否为自身中心的镜像,即判断其是否对称。通过递归和迭代两种方法实现,重点介绍了递归方法中如何比较二叉树的左子树和右子树的镜像关系。
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