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*/10classSolution {11public:12boolisSymmetric(TreeNode*root) {13if(root ==NULL)14returntrue;15if(...
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代码如下: 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*/10classSolution {11public:12vector<int> LPreOrder(TreeNode* tree){//LDR中序遍历13sta...
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) 1...
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 1. 2. 3. 4. 5. But the following is not: 1
题目判断二叉树是对称 Given a binary tree, check whether it is a mirror of itself (ie, symmetric a...
https://leetcode.com/problems/symmetric-tree/ Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree [1,2,2,3,4,4,3] is symmetric: 1 / \ 2 2 / \ / \ 3 4 4 3 ...
Solved symmetric-tree in JS. File(s) Added: 0101-symmetric-tree.js Language(s) Used: JavaScript Submission URL: https://leetcode.com/problems/symmetric-tree/submissions/1346787470/ Create 0101-symmetric-tree.js Verified 3271b86 aakhtar3 reviewed Sep 1, 2024 View reviewed changes javascript/01...
核心思想:递归判断左子树的右子树和右子树的左子树是否相同并且左子树的左子树和右子树的右子树是否相同。原函数只有一个参数,无法实现递归,所以增加一个有两个参数的函数。 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNo...
package leetcode //递归 func isSymmetric1(root *TreeNode) bool { return symmetricHelper(root, root) } func symmetricHelper(left, right *TreeNode) bool { if left == nil && right == nil { return true } if left == nil || right == nil || left.Val != right.Val { ...