116. Populating Next Right Pointers in Each Node

Problem

Given a binary tree

struct TreeLinkNode {
  TreeLinkNode *left;
  TreeLinkNode *right;
  TreeLinkNode *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.

Note:

• You may only use constant extra space.

• Recursive approach is fine, implicit stack space does not count as extra space for this problem.

• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

Example:

Given the following perfect binary tree,

1
   /  \
  2    3
 / \  / \
4  5  6  7

After calling your function, the tree should look like:

1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \  / \
4->5->6->7 -> NULL

Solution (Java)

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return null;
        }
        if (root.left != null && root.right != null) {
            root.left.next = root.right;
        }
        if (root.next != null && root.right != null) {
            root.right.next = root.next.left;
        }
        connect(root.left);
        connect(root.right);
        return root;
    }
}

Solution (Javascript)

/**
 * // 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) {
  let frontier = [root]
  while (frontier.length) {
    const next = []
    frontier.forEach((node) => {
      if (node && node.left) {
        node.left.next = node.right
        const last = next[next.length - 1]
        if (last) {
          last.next = node.left
        }
        next.push(node.left, node.right)
      }
    })
    frontier = next
  }
  return root
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).