【Leetcode】Populating Next Right Pointers in Each Node in JAVA

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

这道题其实就是分清楚next指的对象，然后递归调用就好了~~~

不要用1，2，3来想，用4和5来作为对象来想如何调用安排：

public class PopNextRightPo {

//	struct TreeLinkNode {
//	      TreeLinkNode *left;
//	      TreeLinkNode *right;
//	      TreeLinkNode *next;
//	    }

public void connect(TreeLinkNode root) {
//		if(root==null)	return;
//		if(root.left!=null)	root.left=root.right;
//		if(root.right!=null){
//			if(root.next!=null)	root.right.next=root.next.left;
//		}
//		connect(root.left);//递归调用
//		connect(root.right);//递归调用
if(root == null){
return;
}

if(root.left != null){
root.left.next = root.right;
}
if(root.right != null){
root.right.next = root.next==null ? null : root.next.left;
}

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

大家想想我注释掉的部分为什么不对？