116. Populating Next Right Pointers in Each Node

题目：

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

 

 

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  Tree Depth-first Search  

链接：  http://leetcode.com/problems/populating-next-right-pointers-in-each-node/

题解：

使用DFS。题目给出完全二叉树，所以只要先判断next节点是否为空，接下来判定root的左右子节点是否为空就可以了。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {

    public void connect(TreeLinkNode root) {

        if(root == null)

            return;

        TreeLinkNode node = root.next;

        

        if(node != null){

            if(node.right != null){

                root.right.next = node.left;    

                root.left.next = root.right;    

            }

        } else {

            if(root.right != null){

                root.right.next = null;    

                root.left.next = root.right;    

            }

       }

        

        connect(root.left);

        connect(root.right);

    }

}

 

测试：