[LeetCode] 116. Populating Next Right Pointers in Each Node

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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:

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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:

116. Populating Next Right Pointers in Each Node

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Input: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

Output: {"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

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.

Note:

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

Explanation 1

  1. We can use recursion method here. Since this is a perfect binary tree, if the root node has left node, then it must has right node, so we can check if it has left node first, then connect it with its right node. To connect the right node’s next node, we first check its parent node has next node, if it has, then we can connect the right node’s next to its parent node’s left node.

Solution 1

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/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) return;
        if (root.left != null) {
            root.left.next = root.right;
        }
        if (root.right != null && root.next != null) root.right.next = root.next.left;
        connect(root.left);
        connect(root.right);
    }
}

Explanation 2

  1. We can use a O(1) space iterative method to solve this problem. We create a start pointer that points to the current level’s beginning, a cur pointer that is used to iterate the current level’s node.

Solution 2

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/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */

public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) return;
        TreeLinkNode start = root;
        TreeLinkNode cur = null;
        while (start.left != null) {
            cur = start;
            while (cur != null) {
                cur.left.next = cur.right;
                if (cur.next != null) cur.right.next = cur.next.left;
                cur = cur.next;
            }
            start = start.left;
        }
    }
}