跳转至

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：

1
2
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：

1
2
输入：root = []
输出：[]

示例 3：

1
2
输入：root = [0]
输出：[0]

提示：

  • 树中结点数在范围 [0, 2000] 内
  • -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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
/**
 * 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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
//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;
        }
    }
}