1361. Validate Binary Tree Nodes LeetCode Solution

In this guide, you will get 1361. Validate Binary Tree Nodes LeetCode Solution with the best time and space complexity. The solution to Validate Binary Tree Nodes problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Validate Binary Tree Nodes solution in C++
Validate Binary Tree Nodes solution in Java
Validate Binary Tree Nodes solution in Python
Additional Resources

1361. Validate Binary Tree Nodes LeetCode Solution image

Problem Statement of Validate Binary Tree Nodes

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

1361. Validate Binary Tree Nodes LeetCode Solution in C++

class Solution {
 public:
  bool validateBinaryTreeNodes(int n, vector<int>& leftChild,
                               vector<int>& rightChild) {
    vector<int> inDegrees(n);
    int root = -1;

    // If the in-degree of any node > 1, return false.
    for (const int child : leftChild)
      if (child != -1 && ++inDegrees[child] == 2)
        return false;

    for (const int child : rightChild)
      if (child != -1 && ++inDegrees[child] == 2)
        return false;

    // Find the root (the node with in-degree == 0).
    for (int i = 0; i < n; ++i)
      if (inDegrees[i] == 0)
        if (root == -1)
          root = i;
        else
          return false;  // There're multiple roots.

    // Didn't find the root.
    if (root == -1)
      return false;

    return countNodes(root, leftChild, rightChild) == n;
  }

 private:
  int countNodes(int root, const vector<int>& leftChild,
                 const vector<int>& rightChild) {
    if (root == -1)
      return 0;
    return 1 +  //
           countNodes(leftChild[root], leftChild, rightChild) +
           countNodes(rightChild[root], leftChild, rightChild);
  }
};
/* code provided by PROGIEZ */

1361. Validate Binary Tree Nodes LeetCode Solution in Java

class Solution {
  public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
    int[] inDegrees = new int[n];
    int root = -1;

    // If the in-degree of any node > 1, return false.
    for (final int child : leftChild)
      if (child != -1 && ++inDegrees[child] == 2)
        return false;

    for (final int child : rightChild)
      if (child != -1 && ++inDegrees[child] == 2)
        return false;

    // Find the root (the node with in-degree == 0).
    for (int i = 0; i < n; ++i)
      if (inDegrees[i] == 0)
        if (root == -1)
          root = i;
        else
          return false; // There're multiple roots.

    // Didn't find the root.
    if (root == -1)
      return false;

    return countNodes(root, leftChild, rightChild) == n;
  }

  private int countNodes(int root, int[] leftChild, int[] rightChild) {
    if (root == -1)
      return 0;
    return 1 + //
        countNodes(leftChild[root], leftChild, rightChild) +
        countNodes(rightChild[root], leftChild, rightChild);
  }
}
// code provided by PROGIEZ