给定一个节点个数为n的二叉树，请你把这个二叉树展开为一条单链表。 1.展开后的链表同样是TreeNode，其中right指针指向下一个节点，left节点为空 2.链表的顺序与给定二叉树的先序遍历顺序相同。 3.该题不需要返回链表或者树，请你在原树上面操作，系统会最后检查原树的情况来判断你的代码是否正确 4.该题有O(1)额外空间复杂度的解法，你能实现吗？传入的TreeNode不计入空间复杂度计算 例如： 原二叉树是 展开后是 数据范围：二叉树的节点数满足 ，二叉树节点值满足
示例1
输入
{1,2,3,4,8}
输出
{1,#,2,#,4,#,8,#,3}
示例2
输入
{0}
输出
{0}
加载中...
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 */ public void expandTree (TreeNode root) { // write code here } }
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * }; */ class Solution { public: /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 */ void expandTree(TreeNode* root) { // write code here } };
#coding:utf-8 # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 # # # @param root TreeNode类 # @return Do not return anything, modify root in-place instead. # class Solution: def expandTree(self , root ): # write code here
using System; using System.Collections.Generic; /* public class TreeNode { public int val; public TreeNode left; public TreeNode right; public TreeNode (int x) { val = x; } } */ class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 */ public void expandTree (TreeNode root) { // write code here } }
/* * function TreeNode(x) { * this.val = x; * this.left = null; * this.right = null; * } */ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 * @return void Do not return anything, modify root in-place instead. */ function expandTree( root ) { // write code here } module.exports = { expandTree : expandTree };
<?php /*class TreeNode{ var $val; var $left = NULL; var $right = NULL; function __construct($val){ $this->val = $val; } }*/ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 * @return NULL */ function expandTree( $root ) { // write code here }
# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 # # # @param root TreeNode类 # class Solution: def expandTree(self , root: TreeNode) -> None: # write code here
package main //import "fmt" import . "nc_tools" /* * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 */ func expandTree( root *TreeNode ) { // write code here }
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; * * C语言声明定义全局变量请加上static，防止重复定义 */ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 */ void expandTree(struct TreeNode* root ) { // write code here }
# class TreeNode # attr_accessor :val, :left, :right # def initialize(val, left = nil, right = nil) # @val, @left, @right = val, left, right # end # end # # 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 # # # @param root TreeNode类 # @return {Void} Do not return anything, modify root in-place instead. # class Solution def expandTree(root) # write code here end end
/** * class TreeNode(var `val`: Int) { * var left: TreeNode = null * var right: TreeNode = null * } */ object Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 * @return Do not return anything, modify root in-place instead. */ def expandTree(root: TreeNode): Unit = { // write code here } }
/** * class TreeNode(var `val`: Int) { * var left: TreeNode? = null * var right: TreeNode? = null * } */ object Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 * @return Do not return anything, modify root in-place instead. */ fun expandTree(root: TreeNode?): Unit { // write code here } }
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 */ public void expandTree (TreeNode root) { // write code here } }
/*class TreeNode { * val: number * left: TreeNode | null * right: TreeNode | null * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) { * this.val = (val===undefined ? 0 : val) * this.left = (left===undefined ? null : left) * this.right = (right===undefined ? null : right) * } * } */ /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 * @return Do not return anything, modify root in-place instead. */ export function expandTree(root: TreeNode): void { // write code here }
/** * public class TreeNode { * public var val: Int * public var left: TreeNode? * public var right: TreeNode? * public init(_ val: Int=0, _ left: TreeNode?=nil, _ right: TreeNode?=nil) { * self.val = val * self.left = left * self.right = right * } */ public class Solution { /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 * @return Do not return anything, modify root in-place instead. */ func expandTree ( _ root: TreeNode?) { // write code here } }
/** * #[derive(PartialEq, Eq, Debug, Clone)] * pub struct TreeNode { * pub val: i32, * pub left: Option<Box<TreeNode>>, * pub right: Option<Box<TreeNode>>, * } * * impl TreeNode { * #[inline] * fn new(val: i32) -> Self { * TreeNode { * val: val, * left: None, * right: None, * } * } * } */ struct Solution{ } impl Solution { fn new() -> Self { Solution{} } /** * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可 * * * @param root TreeNode类 * @return {void} Do not return anything, modify root in-place instead. */ pub fn expandTree(&self, root: Option<Box<TreeNode>>) { // write code here } }
{1,2,3,4,8}
{1,#,2,#,4,#,8,#,3}
{0}
{0}