二叉树 Leetcode 101 对称二叉树

Leetcode 101

要点：1.比较的是左右子树是否相同，不是节点的左右孩子；
2.左子树左右中遍历，右子树右左中遍历；
3.灵活递归。

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:// 1.确定递归参数和返回值bool compare(TreeNode* left, TreeNode* right){// 2.确定终止条件if(left == nullptr && right != nullptr) return false;else if(left != nullptr && right == nullptr) return false;else if(left == nullptr && right == nullptr) return true;else if(left->val != right->val) return false;// 3.单次递归逻辑bool out = compare(left->left, right->right);bool in = compare(left->right, right->left);bool isSame = out && in;return isSame;}bool isSymmetric(TreeNode* root) {if(root->left == nullptr && root->right != nullptr){return false;}else if(root->left != nullptr && root->right == nullptr){return false;}return compare(root->left, root->right); }
};