[leedcode 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  7

 

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

/**
 * 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) {
        //注意：
        //1、默认next为空，因此不用讨论最右侧节点的next问题（包括根节点）
        //2、注意例子中5->6，求非最外侧的右节点的next，此时需要结合根节点是否有next进行判断
        if(root==null) return;
        
        if(root.left!=null){
            root.left.next=root.right;
        }
        if(root.right!=null){
            if(root.next!=null){
                root.right.next=root.next.left;
            }
        }
        connect(root.left);
        connect(root.right);
    }
}