114.二叉树展开为链表

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]

提示:

  • 树中结点数在范围 [0, 2000]

  • -100 <= Node.val <= 100

进阶:你可以使用原地算法(O(1) 额外空间)展开这棵树吗?

题解

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public void flatten(TreeNode root) {
//迭代遍历二叉树
if(root==null)return;
TreeNode dummyNode = new TreeNode();
TreeNode curr = dummyNode;
LinkedList<TreeNode> stack = new LinkedList<>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode node = stack.pop();
curr.right = node;
curr.left = null;
curr = curr.right;
if(node.right!=null)
stack.push(node.right);
if(node.left!=null)
stack.push(node.left);
}
root = dummyNode.right;
}
}
//该方法主要分为4步
//1.获取左边节点为
//2.找到左边节点的最右边节点
//3.最右边节点right指向右边节点
//4.将左节点修改为右边节点
//从顶部向下逐步消除左节点:先把right移动到左边,
class Solution {
public void flatten(TreeNode root) {
TreeNode curr = root;
while (curr != null) {
//如果左边节点不为null
if (curr.left != null) {
TreeNode next = curr.left;
TreeNode predecessor = next;
while (predecessor.right != null) {
predecessor = predecessor.right;
}
predecessor.right = curr.right;
curr.left = null;
curr.right = next;
}
curr = curr.right;
}
}
}