Problem Description

In this problem, you are given a 2D integer array named descriptions. Each element of descriptions is a triple [parent_i, child_i, isLeft_i] where:

• parent_i is the value of a parent node in a binary tree.
• child_i is the value of the child node associated with parent_i.
• isLeft_i determines the position of the child_i with respect to parent_i. A value of 1 means child_i is the left child of parent_i, while a value of 0 means child_i is the right child.

Your task is to build the binary tree described by descriptions and return the root node of that binary tree. It's guaranteed that the input descriptions will describe a valid binary tree where each value is unique.

Flowchart Walkthrough

Let's analyze Leetcode Problem 2196, "Create Binary Tree From Descriptions," using the Flowchart. Here’s a breakdown of the decision process based on the structure of the problem:

Is it a graph?

• Yes: The given descriptions indicate parent-child relations, which inherently outline a graph structure, specifically a tree in this case.

Is it a tree?

• Yes: The nature of the problem, creating a binary tree from descriptions, explicitly states that these relationships form a tree.

Is the problem related to directed acyclic graphs (DAGs)?

• Not directly relevant: While a tree is a specific type of DAG, the problem concerns constructing and perhaps traversing the tree itself rather than exploring typical DAG-specific problems such as topological sorting.

Does the problem involve connectivity?

• Not directly: The problem is structured around constructing a tree from given node relationships, implying connectivity by its nature, but the focus is not on identifying or analyzing separate connected components.

Given the problem's requirement to "create" a structure based on connections (child-parent relationships) provided in a list of descriptions, traversal or inspection of each node or connection is necessary. Therefore, the bipartite structure of a binary tree lends itself well to Depth-First Search (DFS), which can effectively be used to traverse and construct the tree. DFS will allow the building and populating of tree nodes efficiently, establishing parent-child relationships linearly from the list of descriptions.

Conclusion: The flowchart leads us to use DFS to construct the binary tree, directly handling the tree structure creation as outlined by the inputs—child-parent pairs—in an orderly and comprehensive manner.

Intuition

The intuition behind solving this problem involves understanding how a binary tree is structured and how the information given in descriptions can be used to construct such a tree. We need to keep track of the nodes and their relationships.

The solution involves two main steps:

1. Create all nodes and store them in a dictionary with their values as keys, so that any node can be accessed instantly when needed.
2. Set up parent-child relationships according to the descriptions. As we connect children to their parents, we also keep track of which nodes have been assigned as children. This is done because the root of the tree will never be assigned as a child.

Finally, after all descriptions are processed, we find the node that has not been assigned as a child. This node will be the root of our binary tree, and hence, is returned as the result.

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

Solution Approach

The implementation of the solution follows these steps:

• First, we use a dictionary g to map the value of each node to the corresponding TreeNode object. We use the defaultdict from the collections module in Python which helps to automatically create a new TreeNode if the value is not already in the dictionary. This is helpful for quickly accessing and storing nodes by their values.

• As we iterate through each description (parent_i, child_i, isLeft_i), we need to check if the parent and child nodes already exist in the dictionary g. If they do not, we create them and store them. This ensures that all nodes are created before we start setting up relationships.

• Then, based on the isLeft_i value, we link the parent node to the child node by setting either the left or right pointer of the parent's TreeNode object to the child's TreeNode object. This sets up the left and right child relationships as described by the input.

• While setting up these relationships, we also keep a vis set which keeps track of all the nodes that have been assigned as a child to some parent node. This set is crucial for identifying the root node later because the root will not be a child of any node.

• Finally, once all descriptions are processed, we iterate through the nodes in the dictionary g. For each node, we check if it is not present in the vis set. The node not present in vis is the one which has never been assigned as a child, indicating it's the root of the tree. We return this root node.

In terms of data structures, this solution uses:

• A defaultdict to construct the nodes dynamically and access them efficiently.
• A set to keep track of all child nodes encountered.
• The binary tree nodes are represented by instances of a class TreeNode, where each node stores a value and two pointers, one for its left child and one for its right child.

This approach follows a pattern often used in tree construction problems, where relationships between nodes are established based on given pairs or tuples and a data structure is required to instantly access any node based on a unique identifier (in this case, the node's value).

Example Walkthrough

Let's walk through a small example to illustrate the solution approach using the following descriptions array:

descriptions = [[20, 15, 1], [20, 17, 0], [50, 20, 1], [50, 80, 0], [80, 19, 1]]

This input describes a binary tree with the following structure:

       50
      /  \
    20    80
   / \    /
 15  17  19

Step by step, we follow the instructions:

1. We initialize a default dictionary g and a set vis as follows:

   • g = defaultdict(TreeNode) which will hold the nodes.
   • vis = set() which will keep track of all child nodes.

2. As we iterate through the descriptions:

   For the first triple [20, 15, 1]:

   • Create nodes for 20 and 15 if they don't exist in g.
   • Since isLeft_i is 1, link 20's left child to 15.
   • Add 15 to the vis set as it is now a known child node.

   For the second triple [20, 17, 0]:

   • Create nodes for 20 and 17 if they don't exist in g.
   • Link 20's right child to 17 because isLeft_i is 0.
   • Add 17 to the vis set.

   For the third triple [50, 20, 1]:

   • Create a node for 50 (20 is already created).
   • Link 50's left child to 20.
   • Add 20 to the vis set.

   For the fourth triple [50, 80, 0]:

   • Create a node for 80.
   • Link 50's right child to 80.
   • Add 80 to the vis set.

   For the last triple [80, 19, 1]:

   • Create a node for 19.
   • Link 80's left child to 19.
   • Add 19 to the vis set.

3. After processing all descriptions, we find the root node:

   • Iterate through g and find a node that is not in the vis set. In this case, it is the node with value 50 because it's the only node that was not added to vis as it was never assigned as a child. Hence, 50 is the root of the tree.

4. Return the root node (The node with value 50), which points to the complete tree structure as described.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code iterates once through the descriptions list, which results in a time complexity of O(N), where N is the number of elements (or sub-descriptions) in the descriptions array. During each iteration, the code performs a constant amount of work, including adding elements to the graph representation g and the visited set vis, as well as linking parent and child nodes. Because these operations are performed in constant time, they do not affect the overall linear time complexity.

Space Complexity

The space complexity of the algorithm is O(N) as well. This holds because the g dictionary and the vis set store each node and its connections from the descriptions. The memory used by g and vis scales linearly with the number of nodes, resulting in a linear space requirement. Additionally, the constructed tree itself has N nodes; however, this does not add to the space complexity as these nodes are already accounted for in g.

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