力扣0114——二叉树展开为链表

[二叉树展开为链表]

难度：中等

题目描述

给你二叉树的根结点 root ，请你将它展开为一个单链表：

• 展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
• 展开后的单链表应该与二叉树 先序遍历 顺序相同。

示例1

输入： root = [1,2,5,3,4,null,6]
输出：[1,null,2,null,3,null,4,null,5,null,6]

示例2

输入： root = []
输出：[]

示例3

输入： root = [0]
输出：[0]

题解

因为和先序遍历相同，所以可以想到利用先序遍历将所有的结点存储到列表中，之后便可以依次添加右子结点

想法代码

using System;
using System.Collections.Generic;public class TreeNode
{public int val;public TreeNode left;public TreeNode right;public TreeNode(int val = 0, TreeNode left = null, TreeNode right = null){this.val = val;this.left = left;this.right = right;}
}public class Solution
{public static void Main(string[] args){TreeNode root = new TreeNode{val = 1,left = new TreeNode{val = 2,left = new TreeNode(3),right = new TreeNode(4)},right = new TreeNode{val = 5,right = new TreeNode(6)}};Solution solution = new Solution();solution.Flatten(root);solution.OnlyReadRightTree(root);}public IList<TreeNode> treeList = new List<TreeNode>();public void Flatten(TreeNode root){Travel(root);int size = treeList.Count;for (int i = 0; i < size; i++){TreeNode tree = treeList[i];tree.left = null;tree.right = i == size - 1 ? null : treeList[i + 1];}}public void Travel(TreeNode root){if (root == null){return;}treeList.Add(root);Travel(root.left);Travel(root.right);}public void OnlyReadRightTree(TreeNode root){if (root == null){return;}OnlyReadRightTree(root.right);Console.Write(root.val);}
}
return;}OnlyReadRightTree(root.right);Console.Write(root.val);}
}