populating-next-right-pointers-in-each-node-ii（二叉树每层用指针链接）

题目

Follow up for problem “Populating Next Right Pointers in Each Node”.

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,

Given the following binary tree,

1

/ \

2 3

/ \ \

4 5 7

After calling your function, the tree should look like:

1 -> NULL

/ \

2 -> 3 -> NULL

/ \ \

4-> 5 -> 7 -> NULL

java实现

import java.util.LinkedList;
import java.util.Queue;
//利用层次遍历的特点，依次遍历每一层，通过嵌套的一层for循环，很巧妙的控制了每一层的输出
public class Solution {
public void connect(TreeLinkNode root) {
Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
if (root == null)
return;
q.add(root);
while (!q.isEmpty()) {
int count = q.size();
for(int i=0;i<count;i++){
TreeLinkNode node = (TreeLinkNode) q.poll();
if(node.left != null)
q.add(node.left);
if(node.right != null)
q.add(node.right);
if(i != count - 1)
node.next = q.peek();
}

}
}
}