Populating Next Right Pointers in Each Node Python Java Leetcode

Given a binary tree

    struct TreeLinkNode {      TreeLinkNode *left;      TreeLinkNode *right;      TreeLinkNode *next;    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set toNULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,

Given the following perfect binary tree,

         1       /  \      2    3     / \  / \    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL       /  \      2 -> 3 -> NULL     / \  / \    4->5->6->7 -> NULL

JAVA:

(easy)

public void connect(TreeLinkNode root) {    if(root == null)        return;    if(root.left != null){        root.left.next = root.right;        if(root.next != null)            root.right.next = root.next.left;    }    connect(root.left);    connect(root.right);}

(hard)

java:class TreeNode{        int val;        TreeNode left=null;        TreeNode right=null;        void TreeNode(int val){                this.val=val;        }}class TreeLinkNode{        int val;        TreeLinkNode left=null;        TreeLinkNode right=null;        TreeLinkNode next=null;        void TreeLinkNode(int val){                this.val=val;        }}public class Solution{        public TreeLinkList reslove(TreeNode root){                List<List<TreeNode>> list=new ArrayList<List<TreeNode>>();                Queue<TreeNode> queue=new LinkedList<TreeNode>();                                if(root==null) return null;                                queue.offer(root);                while(!queue.isEmpty){                        List<TreeNode> subList=new ArrayList<TreeNode>();                        int queuesize=queue.size();                        for(int i=0;i<queuesize;i++){                                if(queue[i].left!=null){                                        queue.offer(queue.peer().left);                                }                                if(queue[i].right!=null){                                        queue.offer(queue.peer().right);                                }                                subList.add(queue.poll());                        }                        list.add(subList);                }                for(int i=0;i<list.length,i++){                        if(list[i].length>1){                                for(int j=1;j<list[i].length;j++){                                        list[j-1].next=list[j];                                }                                list[list[i].length-1].next=null;                        }                        else{                                list[i].next=null;                        }                }                return list[0][0];                        }}

python:

#coding:utf-8class TreeLinkNode:        def __init__(self,val,left,right,next1):                self.val=val                self.left=left                self.right=right                self.next1=next1class Solution:        def resolve(self,root):                if not root:                        return None                cur=root                next=root.left                while cur.left is not None:                        cur.left.next=cur.right                        if cur.next is not None:                                cur.right.next=cur.next.left                                cur=cur.next                        else:                                cur=next                                next=cur.left