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

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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:

/**
 * 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;
}