Problem Description

You are given a number n, which represents the total number of binary tree nodes. Each node is uniquely numbered from 0 to n-1. Along with this, you are provided two lists, leftChild and rightChild, which represent the indices of the left and right children for each node. If a node does not have a left or right child, its corresponding value in the list will be -1. Your task is to determine if all these nodes can form exactly one valid binary tree.

To qualify as a valid binary tree, there must be exactly one root node (a node without a parent), and each node must have exactly one parent, except for the root. Furthermore, there must be no cycles within the tree structure—no node can be an ancestor of itself.

Flowchart Walkthrough

Let's navigate through the decision-making process using the Flowchart to determine the appropriate algorithm for solving LeetCode problem 1361. Validate Binary Tree Nodes. We'll decide step-by-step:

Is it a graph?

• Yes: The problem involves validating if a collection of nodes forms a valid tree, which can be visualized as a graph where each node potentially has connections (edges).

Is it a tree?

• Yes: The entire goal is to verify if the given structure is a valid binary tree.

Therefore, based on the flowchart, after confirming that we are working with a tree, the appropriate algorithm to use is:

DFS (Depth-First Search)

Conclusion: The flowchart indicates that DFS is suitable for this scenario, as we need to traverse each node and ensure there are no cycles, and that all nodes (except the root) have exactly one parent. These properties are efficiently checked using DFS in structural tree validation problems.

Intuition

The approach to solve this problem involves using a union-find data structure, which is perfect for checking cycles and connections between nodes. The primary idea behind union-find is to group nodes into sets. Each set is identified by a representative, which is one of the nodes in the set (initially, each node is its own representative).

The solution follows these steps:

1. Initialization: Create a union-find structure (p) where each node is initially part of its own set, and a visited array (vis) to keep track of whether a node has already been linked as a child to another node.

2. Iterative Check: Go through each node's children (left and right). If a child node is already visited (vis[j] is true), it means that child has more than one parent, which invalidates the tree structure. If the parent of the node is the same as the parent of its child (find(i) == find(j)), it means that a cycle is detected, which again invalidates the tree.

3. Union Operations: If a child has not been visited and it doesn't create a cycle, we perform a union operation. This involves setting the parent of the child's root to be the root of the current node, and marking the child as visited.

4. Validate Tree: Finally, after iterating through all nodes, we check if there is exactly one node left without a parent (one root), which means that n == 1 at the end of the process after decrementing the count when nodes are linked. If n == 1, it indicates that we have exactly one tree and it is valid, so we return true. Any other value indicates an invalid tree.

In summary, the union-find helps efficiently manage the connectivity between nodes while making it easy to check for cycles and multiple parents, ensuring we end up with a valid binary tree if possible.

Learn more about Tree, Depth-First Search, Breadth-First Search, Union Find, Graph and Binary Tree patterns.

Solution Approach

The implementation of the solution revolves around the union-find algorithm, which is a well-known algorithm used in situations where we need to find and manage connected components or sets, especially when the union and find operations are needed frequently and must perform efficiently.

Here is an explanation of each part of the implementation:

1. Initialization: The lists p and vis are initialized. The list p represents the parent of each node, and it's initialized such that each node is its own parent. The list vis is a boolean array initialized to False, indicating that no nodes have been visited initially.

   p = list(range(n))
   vis = [False] * n

2. Find Function: The find function is a recursive function that traverses the p array to find the root of the set that contains node x. If the current node x is not the root (i.e., p[x] != x), it performs path compression by setting p[x] to the result of a recursive call to find(p[x]). Path compression is a technique that flattens the structure of the sets for faster future lookup.

   def find(x: int) -> int:
       if p[x] != x:
           p[x] = find(p[x])
       return p[x]

3. Iterate Over Nodes: We loop over all nodes, and for each node, we loop over its left and right children.

   for i, (a, b) in enumerate(zip(leftChild, rightChild)):

4. Check and Union: Inside the loop, we check if child node j is not -1, which means the current node has a child. There are two important conditions to check for here:

   • If vis[j] is True, it means the child node j is already connected to a parent, indicating an invalid tree because a node has more than one parent.
   • If find(i) == find(j), it means the parent of node i is in the same set as child j, indicating a cycle in the tree, which invalidates the binary tree property.

   If neither condition is met, we perform a union operation by setting the parent of j's root to be i's root, marking j as visited, and decrementing n by 1, because we have linked j under a parent.

   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

5. Validate by Single Root: After the loop, the last validation step is to check whether n == 1. This validation ensures that there is exactly one root node left, implying that the remaining unvisited node is the root of the entire binary tree. If this condition holds, it means all nodes have been connected properly and exactly one valid binary tree has been formed.

   return n == 1

In conclusion, by utilizing the union-find algorithm and taking into account the unique properties of a binary tree, we can determine whether the given nodes can form exactly one valid binary tree.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach provided.

• Suppose we have n = 4 nodes, which are numbered 0 to 3.
• Let's say the leftChild list is [1, -1, -1, -1] and the rightChild list is [2, 3, -1, -1]. This implies that:
  • Node 0 has 1 as its left child and 2 as its right child.
  • Node 1 does not have any children, but it is the right child of Node 0.
  • Node 2 is the left child of Node 0 and does not have any children.
  • Node 3 does not have any children but is a right child of Node 1.

Using the solution approach:

1. Initialization: We initialize parent array p to [0, 1, 2, 3] and visited array vis to [False, False, False, False].

2. Find Function: The find function will be used later to determine the roots of the nodes and apply path compression.

3. Iterate Over Nodes:

   • For Node 0, leftChild[0] = 1 and rightChild[0] = 2.
   • For Node 1, leftChild[1] = -1 (no left child) and rightChild[1] = 3.

4. Check and Union:

   First, iterate for Node 0:

   • For leftChild[0] = 1:
     • vis[1] is False, so Node 1 has no parent yet.
     • find(0) is 0 and find(1) is 1. They are not the same, no cycle here.
     • Perform union: set p[1] to 0 (p[find(1)] = find(0)), mark vis[1] as True, decrement n to 3.
   • For rightChild[0] = 2:
     • vis[2] is False. Node 2 has no parent yet.
     • find(0) is 0 and find(2) is 2. They are not the same, no cycle here.
     • Perform union: set p[2] to 0 (p[find(2)] = find(0)), mark vis[2] as True, decrement n to 2.

Then, iterate for Node 1:

• For rightChild[1] = 3:
  • vis[3] is False. Node 3 has no parent yet.
  • find(1) is 0 and find(3) is 3. They are not the same, no cycle here.
  • Perform union: set p[3] to 0 (p[find(3)] = find(1)), mark vis[3] as True, decrement n to 1.

5. Validate by Single Root: After looping through all nodes, we check if n == 1. Since n has been decremented correctly to 1, we have exactly one node left without a parent. This indicates there is exactly one root in the binary tree, and the tree is valid.

If we visualize the final binary tree formed, it looks like this:

    0
   / \
  1   2
   \
    3

The tree formed is indeed a valid binary tree with Node 0 as the root, and each node has exactly one parent, fulfilling the requirements of the problem. Therefore, the implemented solution would return True.

Solution Implementation

Time and Space Complexity

The given code snippet is a Python method designed to validate a binary tree given the number of nodes n and two lists denoting the indices of left and right children. The function utilizes the Union-Find algorithm, also known as the Disjoint Set Union (DSU) algorithm, to determine if the given input constitutes a valid tree structure.

Time Complexity

The time complexity of this algorithm is primarily determined by two factors: the traversals over the leftChild and rightChild lists, and the union-find operations.

1. Iterating over the leftChild and rightChild lists occurs once for each node in the binary tree. This operation has a linear time complexity of O(n) because each node's children are checked exactly once.

2. The union-find operations consist of the find function, which includes path compression. While a naïve implementation of union-find has a time complexity approaching O(n) for each find operation, path compression optimizes this to an amortized time complexity of approximately O(α(n)), where α(n) is the inverse Ackermann function. The inverse Ackermann function grows extremely slowly and, for all practical purposes, can be considered almost constant.

Considering these factors, the overall time complexity of the union-find operations is O(n * α(n)). However, due to the extremely slow growth of the α(n) term, the time complexity is effectively O(n). Therefore, the total time complexity of the validateBinaryTreeNodes function is O(n).

Space Complexity

The space complexity is determined by the space required to maintain the p array and the vis array, used to track the parents in the union-find structure and the visited state of nodes, respectively.

1. The p array is initialized with a size of n, resulting in a space complexity of O(n).

2. The vis array is also of size n, contributing another O(n) to the space complexity.

These are the only significant sources of space usage in the algorithm. Since they are linear with respect to the number of nodes n, the overall space complexity is also O(n).

In summary, for the provided validateBinaryTreeNodes function:

• The time complexity is O(n).
• The space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.