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问题:判断二叉树是否为镜像二叉树
分析:递归判断,根节点单独判断,然后递归左结点和右结点,之后每次一起递归左结点的左结点和右结点的右结点比较,左结点的右结点和右结点的左结点比较。
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 || (!root->left && !root->right)) 14 return true; 15 return dfs(root->left, root->right); 16 } 17 18 bool dfs(TreeNode *left, TreeNode *right) 19 { 20 if (!left && !right) 21 return true; 22 if ((!left && right) || (left && !right) || (left->val != right->val)) 23 return false; 24 return dfs(left->left, right->right) && dfs(left->right, right->left); 25 } 26 };
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原文地址:http://www.cnblogs.com/bournet/p/4388801.html