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

要将原来的TreeNode增加一个next指针 指向其同层的下一个 很明显地想到是用层序遍历的方法 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 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);
					que.offer(root.right);
					level++;
					level++;
				}
			}
			count=level;
		}
    }
}
原来题中要求的是空间复杂度为0（1） 但遍历的话空间复杂度为O（log N）上面方法不满足要求  从例子来看 每一层的next设置都可以根据自身和上一层的结果来设置 所以采用递归即可 代码如下：

public class Solution {
    public void connect(TreeLinkNode root) {
        if(root==null)return ;
        if(root.left!=null){
            root.left.next=root.right;
        }
        if(root.right!=null&&root.next!=null){
            root.right.next=root.next.left;
        }else{
            if(root.right!=null){
                root.right.next=null;
            }
        }
        connect(root.left);
        connect(root.right);
    }
}


 

Populating Next Right Pointers in Each Node