Populating Next Right Pointers in Each Node II

Problem

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

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

We will implement a BFS solution. This solution is almost identical to the Populating Next Right Pointers in Each Node problem. The only difference is that we need to check if left/right child of node exists before pushing each into queue.

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

We will implement a BFS solution. This solution is almost identical to the Populating Next Right Pointers in Each Node follow-up problem. The only difference is that we need to check if left/right child of node exists before pushing each into queue.