LeetCode //C - 117. Populating Next Right Pointers in Each Node II

117. Populating Next Right Pointers in Each Node II

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 1:

Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
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. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 6000].
  • -100 <= Node.val <= 100

Follow-up:

  • You may only use constant extra space.
  • The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.

From: LeetCode

Link: 117. Populating Next Right Pointers in Each Node II


Solution:

Ideas:

The approach we will follow is:

  1. Traverse the tree level by level (Breadth First Search).
  2. For each level, go through each node and set the next pointer to the next node in that level.
  3. If there's no next node in that level, set the next pointer to NULL.
Code:
c 复制代码
/**
 * Definition for a Node.
 * struct Node {
 *     int val;
 *     struct Node *left;
 *     struct Node *right;
 *     struct Node *next;
 * };
 */

struct Node* connect(struct Node* root) {
    if (!root) return NULL;

    struct Node* prev = NULL;     // Previous node in the current level
    struct Node* head = NULL;     // Head node of the next level
    struct Node* curr = root;     // Current node of the current level

    while (curr) {
        while (curr) {
            // Process the left child
            if (curr->left) {
                if (prev) {
                    prev->next = curr->left;
                } else {
                    head = curr->left;
                }
                prev = curr->left;
            }

            // Process the right child
            if (curr->right) {
                if (prev) {
                    prev->next = curr->right;
                } else {
                    head = curr->right;
                }
                prev = curr->right;
            }

            // Move to the next node in the current level
            curr = curr->next;
        }

        // Move to the next level
        curr = head;
        head = NULL;
        prev = NULL;
    }

    return root;
}
相关推荐
Univin1 分钟前
C++(10.5)
开发语言·c++·算法
Asmalin17 分钟前
【代码随想录day 35】 力扣 01背包问题 一维
算法·leetcode·职场和发展
剪一朵云爱着20 分钟前
力扣2779. 数组的最大美丽值
算法·leetcode·排序算法
qq_4286396124 分钟前
虚幻基础:组件间的联动方式
c++·算法·虚幻
深瞳智检1 小时前
YOLO算法原理详解系列 第002期-YOLOv2 算法原理详解
人工智能·算法·yolo·目标检测·计算机视觉·目标跟踪
tao3556671 小时前
【Python刷力扣hot100】283. Move Zeroes
开发语言·python·leetcode
怎么没有名字注册了啊1 小时前
C++后台进程
java·c++·算法
迎風吹頭髮2 小时前
UNIX下C语言编程与实践32-UNIX 僵死进程:成因、危害与检测方法
服务器·c语言·unix
Rubisco..2 小时前
codeforces 2.0
算法
未知陨落2 小时前
LeetCode:98.颜色分类
算法·leetcode