leetcode116. Populating Next Right Pointers in Each Node

116. 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  7After calling your function, the tree should look like:         1 -> NULL       /  \      2 -> 3 -> NULL     / \  / \    4->5->6->7 -> NULL

解法

递归，从根节点开始，每个左结点的next指向右结点，如果该根节点有next不为null，则该根节点的右结点指向根节点next的左结点。

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