Populating Next Right Pointers in Each Node II

标签：binary   tree   level order   queue   

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;
		}
    }
}

Populating Next Right Pointers in Each Node II

标签：binary   tree   level order   queue   

原文地址：http://blog.csdn.net/u012734829/article/details/42389677