LeetCode 117. Populating Next Right Pointers in Each Node II-优快云博客

本文深入解析了LeetCode上关于填充二叉树节点的进阶问题，提供了迭代法的高效解答策略。通过具体实例展示如何使用常数额外空间解决该问题，并详细阐述了解题思路及实现细节。

原题链接在这里：https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

题目：

Given a binary tree

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *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.

Example:

Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}

Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.

题解：

是Populating Next Right Pointers in Each Node的进阶版.

Iteration解法相同.

Time Complexity: O(n). Space: O(1).

AC Java:

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4     public int val;
 5     public Node left;
 6     public Node right;
 7     public Node next;
 8 
 9     public Node() {}
10 
11     public Node(int _val,Node _left,Node _right,Node _next) {
12         val = _val;
13         left = _left;
14         right = _right;
15         next = _next;
16     }
17 };
18 */
19 class Solution {
20     public Node connect(Node root) {
21         Node cur = root;
22         while(cur != null){
23             Node dummyHead = new Node(); //记录下一层的假头
24             Node it = dummyHead;
25             
26             while(cur != null){
27                 if(cur.left != null){
28                     it.next = cur.left;
29                     it = it.next;
30                 }
31                 
32                 if(cur.right != null){
33                     it.next = cur.right;
34                     it = it.next;
35                 }
36                 
37                 cur = cur.next;  //cur 更新到下一层假头的next上面
38             }
39             
40             cur = dummyHead.next;
41         }
42         
43         return root;
44     }
45 }