#1361

Medium

auto_awesomeGraph traversal with depth-first search

LeetCode Problem Workspace

Validate Binary Tree Nodes

Validate Binary Tree Nodes problem requires checking if a set of nodes forms a valid binary tree using graph traversal.

tree depth first search breadth first search union find graph

Problem Statement

Given n binary tree nodes, each node has two children, a left child and a right child, represented by two arrays leftChild and rightChild. Each element in these arrays corresponds to a node and indicates its left and right children by index. If a node has no child, it is represented as -1. The task is to verify if these nodes form a single valid binary tree. In a valid binary tree, every node must have exactly one parent, and there must be no cycles.

You are to check the structure formed by the arrays and return true if the nodes form a valid tree. If there are multiple roots, cycles, or disconnected components, return false. The binary tree must meet the tree properties, where each node has at most one parent, and the structure must be connected.

Examples

Example 1

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

Output: true

Example details omitted.

Example 2

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

Output: false

Example details omitted.

Example 3

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

Output: false

Example details omitted.

Constraints

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

Solution Approach

Depth-First Search with Parent Tracking

A depth-first search (DFS) can be applied to validate the structure of the binary tree. While traversing, track the parent of each node to ensure each node has only one parent. If any node appears more than once as a child, or if the parent-child relationship is violated, the structure is invalid.

Cycle Detection

Use cycle detection during DFS to ensure that no node is revisited in the traversal. A cycle would indicate that the graph is not a tree, violating the tree properties. You can mark visited nodes and ensure that each node is visited exactly once.

Root Detection

Check for exactly one root node. A valid binary tree should have one node that does not appear as a child of any other node. If there are multiple root nodes, the structure is not a valid tree.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(n)

The time complexity is O(n) due to the single DFS traversal of the nodes, where n is the number of nodes. The space complexity is also O(n), accounting for the storage of visited nodes and parent relationships during traversal.

What Interviewers Usually Probe

Look for an understanding of tree structure properties and cycle detection.
Check if the candidate handles edge cases like multiple roots or nodes with no parents.
Evaluate how efficiently the candidate applies DFS and parent tracking to solve the problem.

Common Pitfalls or Variants

Common pitfalls

Failing to handle cycles correctly in the graph can lead to incorrect results.
Incorrectly identifying the root node, especially in cases where there is more than one root.
Not accounting for the scenario where nodes might not form a connected structure.

Follow-up variants

What if the binary tree is sparse or very large? Can your solution handle it efficiently?
How would you handle the case where some nodes are missing their children?
Can you modify the solution to identify the specific type of error (cycle, multiple roots, etc.)?

FAQ

How can I use DFS to validate a binary tree structure?

By performing a DFS traversal, you can track parent relationships and detect cycles, ensuring that the structure forms a valid tree.

What is the time complexity of the solution for this problem?

The time complexity is O(n), where n is the number of nodes, due to the need to traverse the entire graph once.

Why do we need to detect the root node in this problem?

A valid binary tree must have exactly one root node, which does not appear as a child of any other node. Multiple roots indicate an invalid tree.

What common mistakes should I avoid when solving the 'Validate Binary Tree Nodes' problem?

Avoid missing cycles, incorrectly identifying root nodes, and failing to detect disconnected structures in the graph.

How can GhostInterview assist in solving graph traversal problems like this one?

GhostInterview provides step-by-step assistance in applying DFS, tracking parent-child relationships, and detecting errors in tree structures.

terminal

Solution

Solution 1: Union-Find

We can traverse each node $i$ and its corresponding left and right children $l$, $r$, using an array $vis$ to record whether the node has a parent:

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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

Continue Practicing

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