Create Binary Tree From Descriptions

class Solution {
public:
    TreeNode* constructBinaryTree(vector<vector<int>>& data) {
        unordered_map<int, TreeNode*> nodes;
        unordered_set<int> childNodes;

        for (const auto& item : data) {
            int parent = item[0];
            int child = item[1];
            bool isLeftChild = item[2] == 1;

            if (nodes.find(parent) == nodes.end()) {
                nodes[parent] = new TreeNode(parent);
            }
            if (nodes.find(child) == nodes.end()) {
                nodes[child] = new TreeNode(child);
            }

            if (isLeftChild) {
                nodes[parent]->left = nodes[child];
            } else {
                nodes[parent]->right = nodes[child];
            }

            childNodes.insert(child);
        }

        for (const auto& node : nodes) {
            if (childNodes.find(node.first) == childNodes.end()) {
                return node.second;
            }
        }

        return nullptr;
    }
};

In the given C++ solution, the task is to construct a binary tree from a list of descriptions, where each description specifies the parent-child relationship and whether the child is on the left or right.

Structure of the Solution:

• A class Solution is defined, containing a single public member function constructBinaryTree.
• The function accepts a reference to a 2D vector of integers called data, where each sub-vector contains three elements: [parent, child, isLeftChild].
• Two data structures are employed:
  • unordered_map<int, TreeNode*> nodes to map node values to their respective TreeNode objects.
  • unordered_set<int> childNodes to track which nodes are children.

Steps to Construct the Binary Tree:

1. Iterate over each item in data:
   1. Extract parent and child node values, and determine if the child is a left child.
   2. Create TreeNode objects for the parent and child if they do not yet exist in the nodes map.
   3. Link the child node to its respective parent node based on the isLeftChild flag.
   4. Record every child node in childNodes.
2. After processing all descriptions, search for the root node:
   1. Iterate over nodes. The node that is not in childNodes is the root (since it has no parent).
3. Return the root TreeNode.

This implementation leverages hash tables for fast lookups and insertion, ensuring an efficient construction process. If no root node is found, the function returns nullptr, indicating an invalid input or empty description list. This solution efficiently sets up the binary tree structure as described per input rules.

class Solution {

    public TreeNode constructTree(int[][] descs) {
        // Hold TreeNode references mapped by value
        Map<Integer, TreeNode> nodes = new HashMap<>();

        // Track all child nodes' values
        Set<Integer> childNodes = new HashSet<>();

        // Build and structure the nodes as described
        for (int[] desc : descs) {
            // Parent and child values with child position (left: 1, right: 0)
            int parentVal = desc[0];
            int childVal = desc[1];
            boolean isLeftChild = desc[2] == 1;

            // Ensure both parent and child nodes exist in the map
            if (!nodes.containsKey(parentVal)) {
                nodes.put(parentVal, new TreeNode(parentVal));
            }
            if (!nodes.containsKey(childVal)) {
                nodes.put(childVal, new TreeNode(childVal));
            }

            // Link child to the appropriate side of the parent
            if (isLeftChild) {
                nodes.get(parentVal).left = nodes.get(childVal);
            } else {
                nodes.get(parentVal).right = nodes.get(childVal);
            }

            // Add this node to child nodes tracker
            childNodes.add(childVal);
        }

        // Identify the root node (which is not marked as a child)
        for (TreeNode node : nodes.values()) {
            if (!childNodes.contains(node.val)) {
                return node; // Found the root node
            }
        }

        return null; // Not expected to be reached as per problem constraints
    }
}

This Java solution involves constructing a binary tree from a given set of descriptions where each description specifies a parent-child relationship and whether the child is left or right. Here's an outline of how the provided code works:

• Initialize a Map called nodes to store TreeNode objects referenced by their integer values and a Set called childNodes to keep track of all keys that are children.
• Iterate over the array of descriptions. For each entry, determine the parent and child values as well as the child's position (left or right).
  • If nodes for the parent or child do not exist in the nodes map, create these TreeNode objects and add them to the map.
  • Depending on whether the position is left (desc[2] is 1) or right (desc[2] is 0), link the child node to the appropriate side of the parent node.
  • Add the child value to childNodes to track that this node is a child.
• After building the nodes, determine the root by finding a TreeNode in nodes.values() that is not present in childNodes. This node, which isn't marked as a child, is the root of the tree.
• Return the root node.

The method ensures that each parent and child relationship defined in the input is accurately represented in the tree structure, and it efficiently identifies the root node by checking which node is not listed as a child.

class Solution:
    def createBinaryTree(self, descriptions: List[List[int]]) -> Optional[TreeNode]:
        value_to_node = {}
        child_values = set()

        for desc in descriptions:
            parent_val, child_val, is_left_flag = desc
            is_left = is_left_flag == 1

            if parent_val not in value_to_node:
                value_to_node[parent_val] = TreeNode(parent_val)
            if child_val not in value_to_node:
                value_to_node[child_val] = TreeNode(child_val)

            if is_left:
                value_to_node[parent_val].left = value_to_node[child_val]
            else:
                value_to_node[parent_val].right = value_to_node[child_val]

            child_values.add(child_val)

        for n in value_to_node.values():
            if n.val not in child_values:
                return n

        return None

To build a binary tree from a given set of descriptions using Python3, follow this concise strategy outlined in the code example:

1. Initialize a dictionary value_to_node to map node values to their respective TreeNode instances.

2. Create a set child_values to keep track of all values that appear as children.

3. Loop through each description in the provided list, with each description containing a parent value, a child value, and a flag indicating whether the child is a left or right child:

   • Extract the parent value, child value, and the is_left_flag from the description. Determine if the child is a left node using the flag (is_left = is_left_flag == 1).
   • If the parent value or child value is not already present in value_to_node, create a new TreeNode for it and add it to the dictionary.
   • Depending on the value of is_left, set the left or right child of the parent TreeNode appropriately.
   • Add the child value to the child_values set.

4. After processing all descriptions, determine which node is the root. The root node is the only node that doesn't appear as a child in any description. Iterate over the nodes in value_to_node, and return the one not found in child_values as the root.

If all nodes are accounted as children, the function returns None indicating no isolated root node exists, which usually reflects an inconsistency in the input data.

This approach ensures efficient creation of the tree with clear separation between node creation and tree structure assignments. Make sure to provide the correct input format as a list of descriptions for successful execution.