【二叉树&层次遍历】Populating Next Right Pointers in Each Node

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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

解法：层次遍历，特殊的地方在于遍历完一层之后要添加一个null作为标志，且当队列为空时不能放null否则会死循环

/** * Definition for binary tree with next pointer. * public class TreeLinkNode { *     int val; *     TreeLinkNode left, right, next; *     TreeLinkNode(int x) { val = x; } * } */public class Solution {    public void connect(TreeLinkNode root) {        Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();        if(root == null) return;                q.offer(root);        q.offer(null);        while(!q.isEmpty()){            TreeLinkNode pre = q.poll();            if(pre == null){                if(!q.isEmpty())//判断队列中是否还有元素                q.offer(null);                continue;            }            pre.next = q.peek();            if(pre.left != null) q.offer(pre.left);            if(pre.right != null) q.offer(pre.right);        }    }}

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