【二叉树&层次遍历】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

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);
}
}
}