Populating Next Right Pointers in Each Node II

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

与Populating Next Right Pointers in Each Node 基本一致  只是在向队列中加入子节点时要对左右节点分别判断 代码如下:

public class Solution {
public void connect(TreeLinkNode root) {
if(root==null) return ;
int count=1;
int level=0;
Queue<TreeLinkNode> que =new LinkedList<TreeLinkNode>();
que.offer(root);
while(que.isEmpty()!=true){
level=0;
for(int i=0;i<count;i++){
root=que.peek();
que.poll();
if(i<count-1){
root.next=que.peek();
}else{
root.next=null;
}
if(root.left!=null){
que.offer(root.left);
level++;
}
if(root.right!=null){
que.offer(root.right);
level++;
}
}
count=level;
}
}
}

要求空间负载度为常数  所以上方法不满足要求 与I一样 对上一层节点的next处理完后再处理下一层节点 只是考虑下层节点时 要分别考虑其左右子节点的空否状态 PS：进行递归时 要先右再左 右边节点完成后才能使得左边的next节点得到完善 代码如下：

<pre name="code" class="java">public class Solution {
public void connect(TreeLinkNode root) {
if(root==null)return ;
if(root.left!=null){
if(root.right!=null){
root.left.next=root.right;
}else{
TreeLinkNode p=root.next;
while(p!=null){
if(p.left!=null){
root.left.next=p.left;
break;
}
if(p.right!=null){
root.left.next=p.right;
break;
}
p=p.next;
}
}
}
if(root.right!=null){
TreeLinkNode p=root.next;
while(p!=null){
if(p.left!=null){
root.right.next=p.left;
break;
}
if(p.right!=null){
root.right.next=p.right;
break;
}
p=p.next;
}
}

connect(root.right);
connect(root.left);
}
}