Count minimum cost in binary tree

Question

Given a binary tree where each node has a cost associated with it, write a function to calculate the minimum total cost to reach from the root node to any leaf node. The function should take the root of the binary tree as input and return an integer representing the minimum cost.

Codemia · Accepted Answer

The problem requires us to find the minimum cost to reach any leaf node in a binary tree. The underlying pattern that applies here is Depth-First Search (DFS). Since we are dealing with a tree structure and need to explore all paths from the root to the leaves while accumulating costs, DFS is ideal. It allows us to traverse each path deeply until we reach a leaf, keeping track of the accumulated cost along the way, and then backtrack to explore other paths.

1. Define a recursive function that accepts a node and the current accumulated cost. 
2. If the current node is null, return infinity (or a very large number) since it is not a valid path. 
3. If the current node is a leaf (both left and right children are null), return the accumulated cost of reaching this node. 
4. Recursively call the function for the left child and the right child, adding the current node's cost to the results from the child calls. 
5. Return the minimum cost obtained from the left and right child calls.

**Example:** For a tree structured as follows: 
```
       1
      / \
     2   3
    / \
   4   5
```
The costs are 1, 2, 3, 4, and 5 respectively. The paths are: 
- Path to leaf 4: Cost = 1 + 2 + 4 = 7 
- Path to leaf 5: Cost = 1 + 2 + 5 = 8 
- Path to leaf 3: Cost = 1 + 3 = 4 
The minimum cost is 4.

Here's the Python implementation of the described approach:

```python
class TreeNode:
    def __init__(self, cost=0, left=None, right=None):
        self.cost = cost
        self.left = left
        self.right = right

class Solution:
    def minCost(self, root: TreeNode) -> int:
        if not root:
            return float('inf')  # or a large number
        if not root.left and not root.right:
            return root.cost
        left_cost = self.minCost(root.left)
        right_cost = self.minCost(root.right)
        return root.cost + min(left_cost, right_cost)
```
This function checks each node, accumulates costs, and determines the minimum cost path to any leaf node.

The time complexity of this solution is O(N), where N is the number of nodes in the binary tree. This is because we visit each node exactly once. The space complexity is O(H), where H is the height of the tree, due to the recursive call stack. In the worst case (for a skewed tree), H can be N, leading to a space complexity of O(N). However, for a balanced tree, H would be log(N).

1. If the tree is empty (root is None), return a very large number or some indication of invalid input (e.g., return 0 or raise an exception). 
2. If there is only one node (the root is a leaf), return the cost of that single node. 
3. If all nodes have a negative cost, ensure that the algorithm still correctly identifies the minimum cost. 
4. If there are duplicate costs among nodes, the algorithm should handle them correctly and still return the correct minimum cost.

Count minimum cost in binary tree

DE Shaw

What the Interviewer Expects

Key Topics to Cover

How to Approach This

Possible Follow-up Questions

Sharpen Your Skills on Codemia

Sample Answer

Problem Analysis

Approach

Submit Your Answer

DE Shaw Software Engineer Interview Guide

Related Questions

Binary Search on graph

Count kth largest element in array

Find connected components in string

Merge Three Sorted strings

Binary Search on string