LeetCode #117 — MEDIUM

Populating Next Right Pointers in Each Node II

Move from brute-force thinking to an efficient approach using linked list strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Linked List · Tree

Example 1

[1,2,3,4,5,null,7]

Example 2

[]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #117: Populating Next Right Pointers in Each Node II
/*
// 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 root;
        }
        Deque<Node> q = new ArrayDeque<>();
        q.offerLast(root);
        while (!q.isEmpty()) {
            Node p = null;
            for (int n = q.size(); n > 0; --n) {
                Node node = q.pollFirst();
                if (p != null) {
                    p.next = node;
                }
                p = node;
                if (node.left != null) {
                    q.offerLast(node.left);
                }
                if (node.right != null) {
                    q.offerLast(node.right);
                }
            }
        }
        return root;
    }
}

// Accepted solution for LeetCode #117: Populating Next Right Pointers in Each Node II
/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Left *Node
 *     Right *Node
 *     Next *Node
 * }
 */

func connect(root *Node) *Node {
	if root == nil {
		return root
	}
	q := []*Node{root}
	for len(q) > 0 {
		var p *Node
		for n := len(q); n > 0; n-- {
			node := q[0]
			q = q[1:]
			if p != nil {
				p.Next = node
			}
			p = node
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
	}
	return root
}

# Accepted solution for LeetCode #117: Populating Next Right Pointers in Each Node II
"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""


class Solution:
    def connect(self, root: "Node") -> "Node":
        if root is None:
            return root
        q = deque([root])
        while q:
            p = None
            for _ in range(len(q)):
                node = q.popleft()
                if p:
                    p.next = node
                p = node
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return root

// Accepted solution for LeetCode #117: Populating Next Right Pointers in Each Node II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #117: Populating Next Right Pointers in Each Node II
// /*
// // 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 root;
//         }
//         Deque<Node> q = new ArrayDeque<>();
//         q.offerLast(root);
//         while (!q.isEmpty()) {
//             Node p = null;
//             for (int n = q.size(); n > 0; --n) {
//                 Node node = q.pollFirst();
//                 if (p != null) {
//                     p.next = node;
//                 }
//                 p = node;
//                 if (node.left != null) {
//                     q.offerLast(node.left);
//                 }
//                 if (node.right != null) {
//                     q.offerLast(node.right);
//                 }
//             }
//         }
//         return root;
//     }
// }

// Accepted solution for LeetCode #117: Populating Next Right Pointers in Each Node II
/**
 * Definition for Node.
 * class Node {
 *     val: number
 *     left: Node | null
 *     right: Node | null
 *     next: Node | null
 *     constructor(val?: number, left?: Node, right?: Node, next?: Node) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function connect(root: Node | null): Node | null {
    if (!root) {
        return null;
    }
    const q: Node[] = [root];
    while (q.length) {
        const nq: Node[] = [];
        let p: Node | null = null;
        for (const node of q) {
            if (p) {
                p.next = node;
            }
            p = node;
            const { left, right } = node;
            left && nq.push(left);
            right && nq.push(right);
        }
        q.splice(0, q.length, ...nq);
    }
    return root;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.