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 toNULL.

Initially, all next pointers are set toNULL.

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利用深度遍历是不行了，就从题内看还有没有变得办法，发现给了next，这样就可以用next来连接上层和下层。假设p=root，p==null的话就直接返回，p!=null并且左右子树都存在的话，那么左子树的next应该是右子树。如果p的next存在，p的右子树指向p的next节点的左子树。递归root的左子树和右子树即可。 /** * 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;        }        if(root.left!=null&&root.right!=null){            root.left.next=root.right;        }        if(root.next!=null&&root.right!=null){            root.right.next=root.next.left;        }        connect(root.left);        connect(root.right);    }}