Distribute Coins in Binary Tree

class Solution {
public:
    int coinDistribution(TreeNode* rootNode) {
        totalMoves = 0;
        explore(rootNode);
        return totalMoves;
    }

private:
    int totalMoves; // Store the total number of moves
    int explore(TreeNode* node) {
        if (node == nullptr) return 0;

        int leftExcess = explore(node->left); // excess coins in left child
        int rightExcess = explore(node->right); // excess coins in right child

        totalMoves += abs(leftExcess) + abs(rightExcess); // accumulate the moves needed

        return (node->val - 1) + leftExcess + rightExcess; // calculate current node excess
    }
};

The provided solution is a C++ class designed to determine the number of moves needed to distribute coins in a binary tree such that each node has exactly one coin. Here's a quick breakdown of how this solution operates:

• The class Solution has a private integer member totalMoves which keeps track of the total number of coin redistribution moves required.

• The coinDistribution function is the public interface of the class, taking a pointer to the tree's root node as its argument. The function first sets totalMoves to zero, then calls the explore function starting from the root node and returns the value of totalMoves.

• The private member function explore works recursively to calculate the excess or deficit of coins at each node in the tree:

  • If the current node is nullptr, it returns 0, indicating no excess or deficiency.

  • It recursively calculates the excess coins for both the left and right children of the current node.

  • The totalMoves is incremented by the sum of the absolute values of the left and right excesses. This increment represents the number of moves needed to balance the left and right children with the current node.

  • Finally, it returns the excess coins at the current node, calculated as (node->val - 1) plus the left and right excess values. The -1 accounts for the coin that is to remain with the current node.

This solution efficiently calculates the minimum moves required by balancing coins from the bottom of the tree upwards, ensuring each node ends with exactly one coin after all redistributions.

class Solution {

    private int totalMoves;

    public int distributeCoins(TreeNode node) {
        totalMoves = 0;
        traverse(node);
        return totalMoves;
    }

    private int traverse(TreeNode node) {
        if (node == null) return 0;

        int leftExcess = traverse(node.left);
        int rightExcess = traverse(node.right);

        totalMoves += Math.abs(leftExcess) + Math.abs(rightExcess);

        return node.val - 1 + leftExcess + rightExcess;
    }
}

The problem set involves redistributing coins such that every node of a binary tree has exactly one coin using the minimum number of moves. In this Java solution, a recursion-based approach is used to achieve the objective.

• Start with defining a private variable totalMoves to keep track of the number of moves made during redistribution.
• Create a public function distributeCoins that initializes totalMoves to zero and starts the recursion by calling the traverse function on the tree's root. This function returns the totalMoves counter when completed.
• Implement a private helper function traverse that uses recursion to navigate through the tree. If the node is null, it returns zero, representing no excess coins to move.
• For each node, calculate the excess coins in the left and right child through recursive calls to traverse.
• Adjust totalMoves by adding the absolute value of the excess coins from both left and right children. This sum represents the minimum moves needed to make the current node balanced.
• Return the excess from the current node which is the node value minus one (to make it balanced) added to the excess coins of left and right children combined. This value represents the net excess (positive or negative) that will be returned to the node's parent.

The essence of the solution lies in the recursive traversal and adjustment of the coin balance from the bottom of the tree to the top, ensuring minimum moves by only handling excess coins at each node.

class Solution:
    def coinDistribution(self, root: Optional[TreeNode]) -> int:
        self.total_moves = 0

        def traverse(node):
            if node == None:
                return 0
            
            # Compute the balance of coins in left and right children nodes
            coins_left = traverse(node.left)
            coins_right = traverse(node.right)

            # Update total moves by adding the movements needed for balance
            self.total_moves += abs(coins_left) + abs(coins_right)

            # Calculate the number of spare coins current node has to give or needs
            return (node.val - 1) + coins_left + coins_right

        traverse(root)

        return self.total_moves

The problem Distribute Coins in Binary Tree involves adjusting the number of coins on each node of a binary tree so that every node has exactly one coin. The challenge tasks you with calculating the minimum number of movements required to achieve this balance.

The provided Python solution involves a class Solution with a method coinDistribution that takes root, a TreeNode representing the root of the binary tree. The approach focuses on a depth-first search traversal of the tree to redistribute and count the movements of coins as follows:

• Initialize a variable self.total_moves to count the required movements to balance the coins.

• Define and use a helper function, traverse, which:

  • Returns 0 if the node is None (base case for recursion)
  • Recursively calculates the required movements of coins in both left and right child nodes of the current node.
  • Uses the absolute value of coins required or surplus in the left and right child nodes to update self.total_moves.
  • Determines the net coin balance for the current node and returns it, indicating how many coins need to be moved in or out.

• Call the helper function traverse starting from the root.

• Finally, return self.total_moves which now contains the minimum number of moves required to distribute the coins properly across the tree nodes.

This solution efficiently computes the result by traversing each node exactly once, making it performant with a complexity closely tied to the number of nodes in the tree.