LeetCode 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

内容：将一棵完美二叉树的每一层都串起来。

思路：关键是读懂题意：如果左孩子为空，则右孩子一定为空，所以左孩子为空，则return；如果左孩子不为空，则右孩子一定不为空，那么我们就可以直接连接左右儿子。

/** * 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 || root.left == null) return;            if (root.left != null)     root.left.next = root.right;    if (root.next != null)    root.right.next = root.next.left;        connect(root.left);    connect(root.right);    }}