root.right = invertTree(tmp);returnroot; } } 3.2 Java实现-非递归 publicclassSolution2{publicTreeNodeinvertTree(TreeNode root){if(root ==null) {returnroot; } Deque<TreeNode> q =newArrayDeque<>(); q.push(root);while(!q.isEmpty()) {TreeNodenode=q.pop();TreeNodetmp=node.left; node....
classSolution:# @param {TreeNode} root# @return {TreeNode}definvertTree(self,root):stack=[root]whilelen(stack)>0:node=stack.pop()ifnotnode:continuenode.left,node.right=node.right,node.leftifnode.left:stack.append(node.left)ifnode.right:stack.append(node.right)returnroot So easy,对不对?
1/**2* Definition for a binary tree node.3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { val = x; }8* }9*/10publicclassSolution {11publicTreeNode invertTree(TreeNode root) {12if(root ==null)13returnroot;14TreeNode leftNode =nul...
21 changes: 21 additions & 0 deletions 21 226. Invert Binary Tree.py Original file line numberDiff line numberDiff line change @@ -0,0 +1,21 @@ from typing import Optionalclass TreeNode: def __init__(self, val=0, left=None, right=None):...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */classSolution{public:TreeNode*invertTree(TreeNode*root){if(root==NULL)returnroot;TreeNode*left=root->left...
(int x):val(x),left(NULL),right(NULL){}};classSolution{public:TreeNode*invertTree(TreeNode*root){if(root==NULL)returnNULL;TreeNode*p;p=root->left;root->left=root->right;root->right=p;// 和实现两个变量的交换接近invertTree(root->left);invertTree(root->right);returnroot;}};// ...
226. Invert Binary Tree aliblielite 来自专栏 · leetcode_python_easy # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # recursively,树的问题大都很适合用递归,因为每一个节点都...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; *///思路:二叉树递归 左右结点交换 在递归翻转左右子树classSolution{public:TreeNode*invertTree(TreeNode*root){...
//Non-RecursionclassSolution {public: TreeNode* invertTree(TreeNode*root) {if(!root)returnNULL; queue<TreeNode*>q; q.push(root);while(!q.empty()) { TreeNode*node =q.front(); q.pop(); TreeNode*tmp = node->left; node->left = node->right; ...
226. Invert Binary Tree QuestionEditorial Solution My Submissions Total Accepted: 109341 Total Submissions: 230799 Difficulty: Easy Invert a binary tree. 4 / \ 2 7 / \ / \ 1 3 6 9 to 4 / \ 7 2 / \ / \ 9 6 3 1 Trivia: ...