114. 二叉树展开为链表

原题

https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/description/

思路

先序遍历 + 双指针

复杂度

时间：O(n)
空间：O(n)

Python代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        l = []

        def preOrder(root):
            if not root:
                return 
            l.append(root)
            preOrder(root.left)
            preOrder(root.right)

        preOrder(root)
        for i in range(len(l)-1):
            curr, next = l[i], l[i+1]
            curr.left = None
            curr.right = next 

Go代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func flatten(root *TreeNode) {
	l := []*TreeNode{}
	var preOrder func(*TreeNode)
	preOrder = func(root *TreeNode) {
		if root == nil {
			return
		}
		l = append(l, root)
		preOrder(root.Left)
		preOrder(root.Right)
	}

	preOrder(root)
	for i := 0; i < len(l)-1; i++ {
		curr, next := l[i], l[i+1]
		curr.Left = nil
		curr.Right = next
	}
}