🌵117. Populating Next Right Pointers in Each Node II
https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
js
/**
* // Definition for a Node.
* function Node(val,left,right,next) {
* this.val = val;
* this.left = left;
* this.right = right;
* this.next = next;
* };
*/
/**
* @param {Node} root
* @return {Node}
*/
function connect(root) {
let parent = null
let leftMost = root
let node = null
while (leftMost) {
parent = leftMost
leftMost = null
node = null
while (parent) {
if (!leftMost) {
leftMost = parent.left
}
if (!leftMost) {
leftMost = parent.right
}
if (parent.left) {
if (node) {
node.next = parent.left
}
node = parent.left
}
if (parent.right) {
if (node) {
node.next = parent.right
}
node = parent.right
}
parent = parent.next
}
}
return root
}js
/**
* // Definition for a Node.
* function Node(val, left, right, next) {
* this.val = val === undefined ? null : val;
* this.left = left === undefined ? null : left;
* this.right = right === undefined ? null : right;
* this.next = next === undefined ? null : next;
* };
*/
/**
* @param {Node} root
* @return {Node}
*/
var connect = function(root) {
if (!root) {
return null
}
let current = [root]
let next = []
while (current.length > 0) {
for (let i = 0; i < current.length; i++) {
if (i < current.length - 1) {
current[i].next = current[i + 1]
}
if (current[i].left) {
next.push(current[i].left)
}
if (current[i].right) {
next.push(current[i].right)
}
}
current = next
next = []
}
return root
};py
"""
# Definition for a Node.
class Node(object):
def __init__(self, val, left, right, next):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution(object):
def connect(self, root):
"""
:type root: Node
:rtype: Node
"""
parent = None
leftMost = root
node = None
while leftMost:
parent = leftMost
leftMost = None
node = None
while parent:
if not leftMost:
leftMost = parent.left
if not leftMost:
leftMost = parent.right
if parent.left:
if node:
node.next = parent.left
node = parent.left
if parent.right:
if node:
node.next = parent.right
node = parent.right
parent = parent.next
return rootpy
"""
# Definition for a Node.
class Node(object):
def __init__(self, val=0, left=None, right=None, next=None):
self.val = val
self.left = left
self.right = right
self.next = next
"""
class Solution(object):
def connect(self, root):
"""
:type root: Node
:rtype: Node
"""
if not root:
return None
current, nextRow = [root], []
while current:
for i, node in enumerate(current):
if i < len(current) - 1:
node.next = current[i + 1]
if node.left is not None:
nextRow.append(node.left)
if node.right is not None:
nextRow.append(node.right)
current = nextRow
nextRow = []
return rootgo
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* Next *TreeNode
* }
*/
func connect(root *TreeNode) *TreeNode {
var parent *TreeNode
var leftMost = root
var node *TreeNode
for leftMost != nil {
parent = leftMost
leftMost = nil
node = nil
for parent != nil {
if leftMost != nil {
leftMost = parent.Left
}
if leftMost != nil {
leftMost = parent.Right
}
if parent.Left != nil {
if node != nil {
node.Next = parent.Left
}
node = parent.Left
}
if parent.Right != nil {
if node != nil {
node.Next = parent.Right
}
node = parent.Right
}
parent = parent.Next
}
}
return root
}