(java) 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:就是一个BFS，把每一层的节点向右链接一下
代码如下(已通过leetcode)
public class Solution {

public void connect(TreeLinkNode root) {

List<TreeLinkNode> list=new ArrayList<TreeLinkNode>();

if(root==null) return;

Queue que=new LinkedList<>();

que.add(root);

BFS(list,que);

List<List<TreeLinkNode>> lists=new ArrayList<List<TreeLinkNode>>();

int count=0;

int i=0;

while(count<list.size()) {

List<TreeLinkNode> temp=list.subList((int)Math.pow(2, i)-1,(int)Math.pow(2, i+1)-1);

lists.add(temp);

count+=(int)Math.pow(2, i);

i++;

}

for(int j=0;j<lists.size();j++) {

int k=0;

while(k<lists.get(j).size()-1) {

lists.get(j).get(k).next=lists.get(j).get(k+1);

k++;

}

}

}

private void BFS(List<TreeLinkNode> list,Queue que) {

// TODO Auto-generated method stub

TreeLinkNode temp;

while(!que.isEmpty()) {

temp=(TreeLinkNode) que.remove();

list.add(temp);

if(temp.left!=null) {

que.add(temp.left);

que.add(temp.right);

}

}

}

}