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/**
* 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:
TreeNode* invertTree(TreeNode* root) {
}
};
golang 解法, 执行用时: 0 ms, 内存消耗: 2.1 MB, 提交时间: 2021-07-26 14:01:52
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func invertTree(root *TreeNode) *TreeNode {
if root == nil {
return nil
}
root.Left, root.Right = root.Right, root.Left
if root.Left != nil {
invertTree(root.Left)
}
if root.Right != nil {
invertTree(root.Right)
}
return root
}
php 解法, 执行用时: 8 ms, 内存消耗: 15 MB, 提交时间: 2021-05-10 17:38:46
/**
* Definition for a binary tree node.
* class TreeNode {
* public $val = null;
* public $left = null;
* public $right = null;
* function __construct($val = 0, $left = null, $right = null) {
* $this->val = $val;
* $this->left = $left;
* $this->right = $right;
* }
* }
*/
class Solution {
/**
* @param TreeNode $root
* @return TreeNode
*/
function invertTree(&$root) {
if ( $root == null ) return $root;
list($root->left, $root->right) = [$root->right, $root->left];
if ( $root->left != null )
$this->invertTree($root->left);
if ( $root->right != null )
$this->invertTree($root->right);
return $root;
}
}
php 解法, 执行用时: 8 ms, 内存消耗: 15.1 MB, 提交时间: 2021-05-10 17:38:32
/**
* Definition for a binary tree node.
* class TreeNode {
* public $val = null;
* public $left = null;
* public $right = null;
* function __construct($val = 0, $left = null, $right = null) {
* $this->val = $val;
* $this->left = $left;
* $this->right = $right;
* }
* }
*/
class Solution {
/**
* @param TreeNode $root
* @return TreeNode
*/
function invertTree($root) {
if ( $root == null ) return $root;
list($root->left, $root->right) = [$root->right, $root->left];
if ( $root->left != null )
$this->invertTree($root->left);
if ( $root->right != null )
$this->invertTree($root->right);
return $root;
}
}
javascript 解法, 执行用时: 92 ms, 内存消耗: 37.2 MB, 提交时间: 2020-09-16 21:22:51
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {TreeNode}
*/
var invertTree = function(root) {
if ( root == null ) return null;
var temp = root.left;
root.left = root.right;
root.right = temp;
if ( root.left ) {
invertTree(root.left);
}
if ( root.right ) {
invertTree(root.right);
}
return root;
};
golang 解法, 执行用时: 0 ms, 内存消耗: N/A, 提交时间: 2018-08-27 11:36:16
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func invertTree(root *TreeNode) *TreeNode {
if root == nil {
return nil
}
var temp *TreeNode = root.Left
root.Left = root.Right
root.Right = temp
if root.Left != nil {
invertTree(root.Left)
}
if root.Right != nil {
invertTree(root.Right)
}
return root
}
java 解法, 执行用时: 0 ms, 内存消耗: N/A, 提交时间: 2018-08-27 11:29:54
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode invertTree(TreeNode root) {
if ( root == null )
return root;
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
if ( root.left != null )
invertTree(root.left);
if ( root.right != null )
invertTree(root.right);
return root;
}
}
python3 解法, 执行用时: 40 ms, 内存消耗: N/A, 提交时间: 2018-08-27 11:24:13
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def invertTree(self, root):
"""
:type root: TreeNode
:rtype: TreeNode
"""
if root == None:
return []
root.left, root.right = root.right, root.left
if root.left != None:
self.invertTree(root.left)
if root.right != None:
self.invertTree(root.right)
return root
