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

/**
* 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) {
//利用使用一个队列
Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
int level = 1;//记录需要出队的元素数，
if(root == null ){
return;
}
queue.offer(root);
int count = 0;
TreeLinkNode p = null;
while(queue.size() > 0){
while(count < level - 1){
p = queue.poll();
p.next = queue.peek();
if(p.left != null && p.right != null){
queue.offer(p.left);
queue.offer(p.right);
}
count++;
}
p = queue.poll();
if(p.left != null && p.right != null){
queue.offer(p.left);
queue.offer(p.right);
}
level = level * 2;
count = 0;
}
}
}

题意就是将同层节点连成链，所以想到的是BFS搜索，使用一个队列存储，由于是完全二叉树，因此可以使用level来方便的控制出队的个数，从而也就处理好了层次的问题。
Runtime: 253
ms