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

BFS：但是空间复杂毒不满足常数要求。

public class Solution {    public void connect(TreeLinkNode root) {        if(root==null) return;        if(root.left==null&&root.right==null){            root.next=null;            return;        }        Queue<TreeLinkNode> queue=new LinkedList<>();        int count=1;        int c=1;        queue.add(root.left);        queue.add(root.right);        TreeLinkNode pre=root;        while(!queue.isEmpty()){            TreeLinkNode node=queue.poll();            if(node.left!=null) queue.add(node.left);            if(node.right!=null) queue.add(node.right);            if(c==count){                pre.next=null;                count*=2;                c=1;            }            else{                pre.next=node;                c++;            }            pre=node;        }    }}

改进的方法：

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