leetcode 116. Populating Next Right Pointers in Each Node

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 to NULL.

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

这道题挺简单的。

要求一层一层赋值next，很容易就想到用 BFS queue 做。我是用的 queue 做的，不过大神说这道题可以在 O(1) space 下解决，提示是：你可以利用你创造好的 next links 。

public void connect(TreeLinkNode root) {if(root==null){return;}Queue<TreeLinkNode> queue=new LinkedList<TreeLinkNode>();queue.offer(root);while(!queue.isEmpty()){int size=queue.size();TreeLinkNode tmp=queue.poll();if(tmp.left!=null){queue.offer(tmp.left);queue.offer(tmp.right);}for(int i=1;i<size;i++){TreeLinkNode theNode=queue.poll();tmp.next=theNode;if(theNode.left!=null){queue.offer(theNode.left);queue.offer(theNode.right);}tmp=theNode;}}return;}

有大神这样做：

public class Solution {    public void connect(TreeLinkNode root) {        TreeLinkNode level_start=root;        while(level_start!=null){            TreeLinkNode cur=level_start;            while(cur!=null){                if(cur.left!=null) cur.left.next=cur.right;                if(cur.right!=null && cur.next!=null) cur.right.next=cur.next.left;                               cur=cur.next;            }            level_start=level_start.left;        }    }}

除此之外，还有大神用递归的方法：

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);}