🌵114. Flatten Binary Tree to Linked List
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/
js
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {void} Do not return anything, modify root in-place instead.
*/
function flatten(root) {
if (!root) {
return
}
flatten(root.left)
flatten(root.right)
if (root.left) {
let current = root.left
while (current.right) {
current = current.right
}
current.right = root.right
root.right = root.left
root.left = null
}
}py
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def flatten(self, root):
"""
:type root: TreeNode
:rtype: void Do not return anything, modify root in-place instead.
"""
if root is None:
return
self.flatten(root.left)
self.flatten(root.right)
if root.left:
current = root.left
while current.right:
current = current.right
current.right = root.right
root.right = root.left
root.left = Nonego
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func flatten(root *TreeNode) {
if root == nil {
return
}
flatten(root.Left)
flatten(root.Right)
if root.Left != nil {
current := root.Left
for current.Right != nil {
current = current.Right
}
current.Right = root.Right
root.Right = root.Left
root.Left = nil
}
}