LeetCode #2196 — MEDIUM

Create Binary Tree From Descriptions

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

The Problem

Problem Statement

You are given a 2D integer array descriptions where descriptions[i] = [parent_i, child_i, isLeft_i] indicates that parent_i is the parent of child_i in a binary tree of unique values. Furthermore,

If isLeft_i == 1, then child_i is the left child of parent_i.
If isLeft_i == 0, then child_i is the right child of parent_i.

Construct the binary tree described by descriptions and return its root.

The test cases will be generated such that the binary tree is valid.

Example 1:

Input: descriptions = [[20,15,1],[20,17,0],[50,20,1],[50,80,0],[80,19,1]]
Output: [50,20,80,15,17,19]
Explanation: The root node is the node with value 50 since it has no parent.
The resulting binary tree is shown in the diagram.

Example 2:

Input: descriptions = [[1,2,1],[2,3,0],[3,4,1]]
Output: [1,2,null,null,3,4]
Explanation: The root node is the node with value 1 since it has no parent.
The resulting binary tree is shown in the diagram.

Constraints:

1 <= descriptions.length <= 10⁴
descriptions[i].length == 3
1 <= parent_i, child_i <= 10⁵
0 <= isLeft_i <= 1
The binary tree described by descriptions is valid.

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array descriptions where descriptions[i] = [parenti, childi, isLefti] indicates that parenti is the parent of childi in a binary tree of unique values. Furthermore, If isLefti == 1, then childi is the left child of parenti. If isLefti == 0, then childi is the right child of parenti. Construct the binary tree described by descriptions and return its root. The test cases will be generated such that the binary tree is valid.

Baseline thinking

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

Pattern signal: Array · Hash Map · Tree

Example 1

[[20,15,1],[20,17,0],[50,20,1],[50,80,0],[80,19,1]]

Example 2

[[1,2,1],[2,3,0],[3,4,1]]

Core Insight

What unlocks the optimal approach

Could you represent and store the descriptions more efficiently?
Could you find the root node?
The node that is not a child in any of the descriptions is the root node.

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 #2196: Create Binary Tree From Descriptions
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode createBinaryTree(int[][] descriptions) {
        Map<Integer, TreeNode> nodes = new HashMap<>();
        Set<Integer> children = new HashSet<>();
        for (var d : descriptions) {
            int parent = d[0], child = d[1], isLeft = d[2];
            if (!nodes.containsKey(parent)) {
                nodes.put(parent, new TreeNode(parent));
            }
            if (!nodes.containsKey(child)) {
                nodes.put(child, new TreeNode(child));
            }
            if (isLeft == 1) {
                nodes.get(parent).left = nodes.get(child);
            } else {
                nodes.get(parent).right = nodes.get(child);
            }
            children.add(child);
        }
        for (var e : nodes.entrySet()) {
            if (!children.contains(e.getKey())) {
                return e.getValue();
            }
        }
        return null;
    }
}

// Accepted solution for LeetCode #2196: Create Binary Tree From Descriptions
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func createBinaryTree(descriptions [][]int) *TreeNode {
	nodes := map[int]*TreeNode{}
	children := map[int]bool{}
	for _, d := range descriptions {
		parent, child, isLeft := d[0], d[1], d[2]
		if _, ok := nodes[parent]; !ok {
			nodes[parent] = &TreeNode{Val: parent}
		}
		if _, ok := nodes[child]; !ok {
			nodes[child] = &TreeNode{Val: child}
		}
		if isLeft == 1 {
			nodes[parent].Left = nodes[child]
		} else {
			nodes[parent].Right = nodes[child]
		}
		children[child] = true
	}
	for k, v := range nodes {
		if _, ok := children[k]; !ok {
			return v
		}
	}
	return nil
}

# Accepted solution for LeetCode #2196: Create Binary Tree From Descriptions
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def createBinaryTree(self, descriptions: List[List[int]]) -> Optional[TreeNode]:
        nodes = defaultdict(TreeNode)
        children = set()
        for parent, child, isLeft in descriptions:
            if parent not in nodes:
                nodes[parent] = TreeNode(parent)
            if child not in nodes:
                nodes[child] = TreeNode(child)
            children.add(child)
            if isLeft:
                nodes[parent].left = nodes[child]
            else:
                nodes[parent].right = nodes[child]
        root = (set(nodes.keys()) - children).pop()
        return nodes[root]

// Accepted solution for LeetCode #2196: Create Binary Tree From Descriptions
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::{HashMap, HashSet};
use std::rc::Rc;
impl Solution {
    pub fn create_binary_tree(descriptions: Vec<Vec<i32>>) -> Option<Rc<RefCell<TreeNode>>> {
        let mut nodes = HashMap::new();
        let mut children = HashSet::new();

        for d in descriptions {
            let parent = d[0];
            let child = d[1];
            let is_left = d[2];

            nodes
                .entry(parent)
                .or_insert_with(|| Rc::new(RefCell::new(TreeNode::new(parent))));
            nodes
                .entry(child)
                .or_insert_with(|| Rc::new(RefCell::new(TreeNode::new(child))));

            if is_left == 1 {
                nodes.get(&parent).unwrap().borrow_mut().left =
                    Some(Rc::clone(nodes.get(&child).unwrap()));
            } else {
                nodes.get(&parent).unwrap().borrow_mut().right =
                    Some(Rc::clone(nodes.get(&child).unwrap()));
            }

            children.insert(child);
        }

        for (key, node) in &nodes {
            if !children.contains(key) {
                return Some(Rc::clone(node));
            }
        }

        None
    }
}

// Accepted solution for LeetCode #2196: Create Binary Tree From Descriptions
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function createBinaryTree(descriptions: number[][]): TreeNode | null {
    const nodes: Record<number, TreeNode> = {};
    const children = new Set<number>();
    for (const [parent, child, isLeft] of descriptions) {
        if (!nodes[parent]) {
            nodes[parent] = new TreeNode(parent);
        }
        if (!nodes[child]) {
            nodes[child] = new TreeNode(child);
        }
        if (isLeft) {
            nodes[parent].left = nodes[child];
        } else {
            nodes[parent].right = nodes[child];
        }
        children.add(child);
    }
    for (const [k, v] of Object.entries(nodes)) {
        if (!children.has(+k)) {
            return v;
        }
    }
}

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

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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.