[LeeCode]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

思路：采用层次遍历的方法，针对每一层进行处理
代码如下：

/**
* 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) {
if (root == null) return;
int thislevel = 1;
int nextlevel = 0;
LinkedList<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
q.add(root);
while(q.isEmpty() == false) {
while(thislevel > 0) {
TreeLinkNode temp = q.poll();
thislevel--;
if (thislevel == 0) temp.next = null;
else temp.next = q.peek();
if(temp.left != null) {
nextlevel += 2;
q.offer(temp.left);
q.offer(temp.right);
}
}
thislevel = nextlevel;
nextlevel = 0;
}

}

}