How do you solve Validate Binary Tree Nodes on LeetCode?

Validate Binary Tree Nodes (LeetCode #1361) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: A valid binary tree must have exactly one root node.

What is the brute force solution for Validate Binary Tree Nodes?

The brute force approach for Validate Binary Tree Nodes is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we check each node to see if it has a parent and count the number of parents for each node. If any node has more than one parent or if there are multiple root candidates, we can conclude that the nodes do not form a valid binary tree. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Validate Binary Tree Nodes?

Clearly explain your thought process and the properties of a binary tree. Consider edge cases, such as when n is small (e.g., 1 or 2). Practice writing clean and efficient code, as clarity is key in interviews.

What are common mistakes when solving Validate Binary Tree Nodes?

Failing to check if a node has more than one parent. Not accounting for the case where multiple nodes could be roots.

#1361

Validate Binary Tree Nodes

Medium

TreeDepth-First SearchBreadth-First SearchUnion-FindGraph TheoryBinary TreeGraph TheoryTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses a single pass to track parent-child relationships and ensures that there is exactly one root node and that no node has more than one parent. This is efficient and avoids unnecessary checks.

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Algorithm

4 steps

1Step 1: Create an array to track the number of parents for each node.
2Step 2: Iterate through each node and for each child (left and right), increment the parent's count.
3Step 3: Count the number of nodes with zero parents (potential roots).
4Step 4: Check that there is exactly one root and that no node has more than one parent.

solution.py11 lines

1# Full working Python code
2
3def validateBinaryTreeNodes(n, leftChild, rightChild):
4    parent_count = [0] * n
5    for i in range(n):
6        if leftChild[i] != -1:
7            parent_count[leftChild[i]] += 1
8        if rightChild[i] != -1:
9            parent_count[rightChild[i]] += 1
10    root_count = sum(1 for count in parent_count if count == 0)
11    return root_count == 1 and all(count <= 1 for count in parent_count)

ℹ

Complexity note: The time complexity is O(n) as we make a single pass through the nodes to count parents, and the space complexity is O(n) due to the parent count array.

1A valid binary tree must have exactly one root node.
2Each node can have at most one parent; no cycles are allowed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.