LeetCode #1361 — MEDIUM

Validate Binary Tree Nodes

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

The Problem

Problem Statement

You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree.

If node i has no left child then leftChild[i] will equal -1, similarly for the right child.

Note that the nodes have no values and that we only use the node numbers in this problem.

Example 1:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
Output: true

Example 2:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]
Output: false

Example 3:

Input: n = 2, leftChild = [1,0], rightChild = [-1,-1]
Output: false

Constraints:

n == leftChild.length == rightChild.length
1 <= n <= 10⁴
-1 <= leftChild[i], rightChild[i] <= n - 1

Step 01

Brute Force Baseline

Problem summary: You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree. If node i has no left child then leftChild[i] will equal -1, similarly for the right child. Note that the nodes have no values and that we only use the node numbers in this problem.

Baseline thinking

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

Pattern signal: Tree · Union-Find

Example 1

4
[1,-1,3,-1]
[2,-1,-1,-1]

Example 2

4
[1,-1,3,-1]
[2,3,-1,-1]

Example 3

2
[1,0]
[-1,-1]

Step 02

Core Insight

What unlocks the optimal approach

Find the parent of each node.
A valid tree must have nodes with only one parent and exactly one node with no parent.

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 #1361: Validate Binary Tree Nodes
class Solution {
    private int[] p;

    public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        boolean[] vis = new boolean[n];
        for (int i = 0, m = n; i < m; ++i) {
            for (int j : new int[] {leftChild[i], rightChild[i]}) {
                if (j != -1) {
                    if (vis[j] || find(i) == find(j)) {
                        return false;
                    }
                    p[find(i)] = find(j);
                    vis[j] = true;
                    --n;
                }
            }
        }
        return n == 1;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

// Accepted solution for LeetCode #1361: Validate Binary Tree Nodes
func validateBinaryTreeNodes(n int, leftChild []int, rightChild []int) bool {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	vis := make([]bool, n)
	for i, a := range leftChild {
		for _, j := range []int{a, rightChild[i]} {
			if j != -1 {
				if vis[j] || find(i) == find(j) {
					return false
				}
				p[find(i)] = find(j)
				vis[j] = true
				n--
			}
		}
	}
	return n == 1
}

# Accepted solution for LeetCode #1361: Validate Binary Tree Nodes
class Solution:
    def validateBinaryTreeNodes(
        self, n: int, leftChild: List[int], rightChild: List[int]
    ) -> bool:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(n))
        vis = [False] * n
        for i, (a, b) in enumerate(zip(leftChild, rightChild)):
            for j in (a, b):
                if j != -1:
                    if vis[j] or find(i) == find(j):
                        return False
                    p[find(i)] = find(j)
                    vis[j] = True
                    n -= 1
        return n == 1

// Accepted solution for LeetCode #1361: Validate Binary Tree Nodes
struct Solution;

impl Solution {
    fn validate_binary_tree_nodes(n: i32, left_child: Vec<i32>, right_child: Vec<i32>) -> bool {
        let n = n as usize;
        let mut indegree = vec![0; n];
        let mut outdegree = vec![0; n];
        let mut edge = 0;
        for i in 0..n {
            if left_child[i] != -1 {
                edge += 1;
                outdegree[i] += 1;
                indegree[left_child[i] as usize] += 1;
            }
            if right_child[i] != -1 {
                edge += 1;
                outdegree[i] += 1;
                indegree[right_child[i] as usize] += 1;
            }
        }
        let degree = (0..n).any(|i| {
            let a = n != 1 && indegree[i] == 0 && outdegree[i] == 0;
            let b = indegree[i] > 1;
            let c = outdegree[i] > 2;
            a || b || c
        });
        !degree && edge + 1 == n
    }
}

#[test]
fn test() {
    let n = 4;
    let left_child = vec![1, -1, 3, -1];
    let right_child = vec![2, -1, -1, -1];
    let res = true;
    assert_eq!(
        Solution::validate_binary_tree_nodes(n, left_child, right_child),
        res
    );
    let n = 4;
    let left_child = vec![1, -1, 3, -1];
    let right_child = vec![2, 3, -1, -1];
    let res = false;
    assert_eq!(
        Solution::validate_binary_tree_nodes(n, left_child, right_child),
        res
    );
    let n = 2;
    let left_child = vec![1, 0];
    let right_child = vec![-1, -1];
    let res = false;
    assert_eq!(
        Solution::validate_binary_tree_nodes(n, left_child, right_child),
        res
    );
    let n = 6;
    let left_child = vec![1, -1, -1, 4, -1, -1];
    let right_child = vec![2, -1, -1, 5, -1, -1];
    let res = false;
    assert_eq!(
        Solution::validate_binary_tree_nodes(n, left_child, right_child),
        res
    );
}

// Accepted solution for LeetCode #1361: Validate Binary Tree Nodes
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1361: Validate Binary Tree Nodes
// class Solution {
//     private int[] p;
// 
//     public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
//         p = new int[n];
//         for (int i = 0; i < n; ++i) {
//             p[i] = i;
//         }
//         boolean[] vis = new boolean[n];
//         for (int i = 0, m = n; i < m; ++i) {
//             for (int j : new int[] {leftChild[i], rightChild[i]}) {
//                 if (j != -1) {
//                     if (vis[j] || find(i) == find(j)) {
//                         return false;
//                     }
//                     p[find(i)] = find(j);
//                     vis[j] = true;
//                     --n;
//                 }
//             }
//         }
//         return n == 1;
//     }
// 
//     private int find(int x) {
//         if (p[x] != x) {
//             p[x] = find(p[x]);
//         }
//         return p[x];
//     }
// }

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 × \alpha(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.

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.