Populating Next Right Pointers in Each Node

Problem

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

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 = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect 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, 212 - 1].
• -1000 <= Node.val <= 1000

Solution

/**
 * // Definition for a Node.
 * function Node(val, left, right, next) {
 *    this.val = val === undefined ? null : val;
 *    this.left = left === undefined ? null : left;
 *    this.right = right === undefined ? null : right;
 *    this.next = next === undefined ? null : next;
 * };
 */

/**
 * @param {Node} root
 * @return {Node}
 */
var connect = function(root) {
    const queue = root ? [root] : [];

    while (queue.length) {
        const n = queue.length;
        for (let i = 0; i < n; i++) {
            const node = queue.shift();
            node.next = i < n - 1 ? queue[0] : null;

            if (node.left) {
                queue.push(node.left);
                queue.push(node.right);
            }
        }
    }
    return root;
};

We will implement a BFS solution. We do a level order traversal to connect the nodes at each level.

Follow-up

You may only use constant extra space.

Solution

/**
 * // Definition for a Node.
 * function Node(val, left, right, next) {
 *    this.val = val === undefined ? null : val;
 *    this.left = left === undefined ? null : left;
 *    this.right = right === undefined ? null : right;
 *    this.next = next === undefined ? null : next;
 * };
 */

/**
 * @param {Node} root
 * @return {Node}
 */
var connect = function(root) {
    const queue = new Node(); // fake queue
    queue.next = root;

    while (queue.next) {
        let size = 0;
        const next = new Node(); // linked list for next level of nodes
        let nextTail = next; // tail of next

        while (queue.next) {
            const node = queue.next;
            queue.next = queue.next.next;

            if (node.left) {
                nextTail.next = node.left;
                nextTail = nextTail.next;
                nextTail.next = node.right;
                nextTail = nextTail.next;
            }
        }
        queue.next = next.next;
    }
    return root;
};

We will implement a BFS solution. Instead of allocating an array as the queue, we use the nodes themselves as a linked list to achieve O(1) extra space.