[LeetCode] 114. Flattern Binary Tree to Linked List（将二叉树扁平化成单链表）

• Difficulty: Medium

• Related Topics: Tree, Depth-first Search

• Link: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/

Description

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

    1
   / 
  2   5
 /    
3   4   6

The flattened tree should look like:

1
 
  2
   
    3
     
      4
       
        5
         
          6

Hints

1. If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.

Solution

题目的意思是：给定一棵二叉树，将其转换为单链表，单链表的顺序为树的前序遍历顺序，right 为单链表的 next。一样地，对于二叉树问题，需要考虑能不能将其分解为子问题。就本题而言，需要做以下步骤：

1. 分别扁平化左子树和右子树

2. 把左子树接到右子树上、右子树接在原左子树的末尾

代码如下：

class Solution {
    fun flatten(root: TreeNode?): Unit {
        if (root == null) {
            return
        }
        val left = root.left
        val right = root.right
        // 扁平化左右子树
        flatten(root.left)
        flatten(root.right)
        // 清除左子树
        root.left = null
        // 将左子树接在原来右子树的位置上
        root.right = left

        // 右子树则接在现在右子树的末尾
        var p = root
        while (p?.right != null) {
            p = p.right
        }
        p?.right = right
    }
}