#979

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Distribute Coins in Binary Tree

Minimize the number of moves needed to distribute coins in a binary tree so that each node has exactly one coin.

Problem Statement

You are given the root of a binary tree with n nodes, where each node has a certain number of coins. The total number of coins in the entire tree is exactly n. In each move, you can move one coin between two adjacent nodes, either from parent to child or from child to parent. Your task is to return the minimum number of moves required to make each node in the tree have exactly one coin.

To achieve this, you will need to calculate the minimum number of coin transfers, traversing the tree and adjusting coin distributions using the fewest possible moves. Make sure that after the moves, each node has exactly one coin, and you track this process efficiently using a binary-tree traversal strategy.

Examples

Example 1

Input: root = [3,0,0]

Output: 2

From the root of the tree, we move one coin to its left child, and one coin to its right child.

Example 2

Input: root = [0,3,0]

Output: 3

From the left child of the root, we move two coins to the root [taking two moves]. Then, we move one coin from the root of the tree to the right child.

Constraints

The number of nodes in the tree is n.
1 <= n <= 100
0 <= Node.val <= n
The sum of all Node.val is n.

Solution Approach

DFS Traversal with State Tracking

Use a depth-first search (DFS) traversal to explore the tree. Track the surplus or deficit of coins at each node during traversal. At each step, calculate the number of coins that need to be moved from a node to its parent or child.

Propagate Coin Movement

As you move through the tree using DFS, propagate the number of coins that need to be transferred up or down the tree. Each move of a coin between adjacent nodes adds to the total number of moves required. This way, you ensure that the total moves reflect the necessary adjustments.

Optimization with Greedy Approach

The problem can be optimized by only counting the total moves needed during the DFS traversal. Since each move involves transferring one coin at a time, you don't need to perform unnecessary recalculations. The greedy approach ensures that you minimize redundant moves and achieve the result in linear time.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(n)

The time complexity of this approach is O(n), where n is the number of nodes in the tree, because we visit each node exactly once. The space complexity is O(n) due to the recursive DFS call stack and the space used for storing the tree's node values.

What Interviewers Usually Probe

Understanding DFS traversal and how to track state across nodes is crucial.
Candidates should be able to explain the greedy approach used to optimize the number of moves.
The ability to optimize space complexity while traversing the tree is a key aspect to evaluate.

Common Pitfalls or Variants

Common pitfalls

Overcomplicating the tree traversal, leading to unnecessary computations.
Forgetting to account for coin transfers between child and parent nodes correctly.
Not properly tracking the surplus or deficit of coins at each node during the DFS traversal.

Follow-up variants

Allowing more than one coin per node, but still minimizing the number of moves to distribute coins evenly.
Considering a non-binary tree structure for a similar distribution problem.
Handling large tree sizes and ensuring the DFS is optimized for performance.

FAQ

How do I approach solving the Distribute Coins in Binary Tree problem?

Use depth-first search (DFS) to traverse the tree and track the number of coins that need to be moved, calculating the minimum moves as you go.

What is the optimal time complexity for solving this problem?

The optimal time complexity is O(n), where n is the number of nodes in the tree, as each node must be visited exactly once.

What are the common pitfalls when solving this problem?

Overcomplicating the traversal or failing to track coin movement correctly between nodes are common pitfalls.

How does DFS help in solving the Distribute Coins in Binary Tree problem?

DFS allows you to efficiently traverse the tree while keeping track of coin surpluses or deficits, ensuring the minimum number of moves.

What is the space complexity of the solution?

The space complexity is O(n) due to the recursion stack during DFS traversal and the need to store node values.

terminal

Solution

Solution 1: DFS

We define a function $\textit{dfs(node)}$, which represents the coin overload in the subtree rooted at $\textit{node}$, i.e., the number of coins minus the number of nodes. If $\textit{dfs(node)}$ is positive, it means the subtree has more coins than nodes, and the excess coins need to be moved out of the subtree; if $\textit{dfs(node)}$ is negative, it means the subtree has fewer coins than nodes, and the shortfall needs to be moved into the subtree.

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def distributeCoins(self, root: Optional[TreeNode]) -> int:
        def dfs(root):
            if root is None:
                return 0
            left, right = dfs(root.left), dfs(root.right)
            nonlocal ans
            ans += abs(left) + abs(right)
            return left + right + root.val - 1

        ans = 0
        dfs(root)
        return ans

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