How do you solve Make Costs of Paths Equal in a Binary Tree on LeetCode?

Make Costs of Paths Equal in a Binary Tree (LeetCode #2673) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with Unknown time complexity and Unknown space complexity. Key insight: Instead of calculating the cost for each path separately, we can directly find the maximum cost path and calculate the increments needed for all other paths to match this maximum. This reduces unnecessary calculations.

What is the time complexity of Make Costs of Paths Equal in a Binary Tree?

The optimal solution for Make Costs of Paths Equal in a Binary Tree runs in Unknown time and uses Unknown space.

What are the interview tips for Make Costs of Paths Equal in a Binary Tree?

Clarify the structure of the binary tree and how nodes are connected. Discuss the importance of identifying the maximum path cost early. Practice tree traversal techniques, as they are commonly used in tree problems.

What are common mistakes when solving Make Costs of Paths Equal in a Binary Tree?

Not considering all leaf nodes when calculating increments. Overcomplicating the traversal logic.

Make Costs of Paths Equal in a Binary Tree — LeetCode #2673 (Medium)

Tags: Array, Dynamic Programming, Greedy, Tree, Binary Tree

Related patterns: Tree Traversal, Dynamic Programming

Brute Force approach

Time complexity: O(n²). Space complexity: O(n).

We can calculate the cost of each path from the root to every leaf node and then find the maximum path cost. We then increment the costs of the other paths to match this maximum cost. This approach is straightforward but inefficient.

The time complexity is O(n²) because for each leaf node, we traverse the entire list of costs to find the maximum cost, which can be inefficient for large trees. The space complexity is O(n) due to the storage of path costs.

Step 1: Traverse the tree to find the cost of each path from the root to every leaf node.
Step 2: Identify the maximum path cost among all leaf nodes.
Step 3: For each leaf node, calculate the increments needed to match the maximum path cost and sum these increments.

1. Start at root (1), current cost = 1. 2. Move to left child (2), current cost = 1 + 5 = 6. 3. Move to left child (4), leaf node, costs = [6 + 2 = 8]. 4. Backtrack to root, move to right child (3), current cost = 1 + 5 = 6. 5. Move to left child (6), leaf node, costs = [8, 6 + 3 = 9]. 6. Move to right child (7), leaf node, costs = [8, 9, 6 + 1 = 7].

Optimal Solution approach

Time complexity: Unknown. Space complexity: Unknown.

Instead of calculating the cost for each path separately, we can directly find the maximum cost path and calculate the increments needed for all other paths to match this maximum. This reduces unnecessary calculations.

Step 1: Traverse the tree and calculate the cost of each path recursively, while keeping track of the maximum path cost.
Step 2: For each leaf node, calculate the increments needed to match the maximum path cost.
Step 3: Return the total increments.

Common Mistakes

Not considering all leaf nodes when calculating increments.
Overcomplicating the traversal logic.

Interview Tips

Clarify the structure of the binary tree and how nodes are connected.
Discuss the importance of identifying the maximum path cost early.
Practice tree traversal techniques, as they are commonly used in tree problems.

#2673

Make Costs of Paths Equal in a Binary Tree

Medium

Array↗Dynamic Programming↗Greedy↗Tree↗Binary Tree↗Tree Traversal↗Dynamic Programming↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	Unknown
Space	O(n)	Unknown

💡

Intuition

Time UnknownSpace Unknown

⚙️

Algorithm

3 steps

1Step 1: Traverse the tree and calculate the cost of each path recursively, while keeping track of the maximum path cost.
2Step 2: For each leaf node, calculate the increments needed to match the maximum path cost.
3Step 3: Return the total increments.

solution.py37 lines

1class TreeNode:
2    def __init__(self, cost):
3        self.cost = cost
4        self.left = None
5        self.right = None
6
7def build_tree(cost):
8    nodes = [TreeNode(c) for c in cost]
9    for i in range(len(nodes)):
10        if 2 * i + 1 < len(nodes):
11            nodes[i].left = nodes[2 * i + 1]
12        if 2 * i + 2 < len(nodes):
13            nodes[i].right = nodes[2 * i + 2]
14    return nodes[0]
15
16def min_increments(n, cost):
17    root = build_tree(cost)
18    max_cost = 0
19    increments = 0
20
21    def dfs(node, current_cost):
22        nonlocal max_cost, increments
23        if not node:
24            return
25        current_cost += node.cost
26        if not node.left and not node.right:
27            increments += max_cost - current_cost
28            max_cost = max(max_cost, current_cost)
29        else:
30            dfs(node.left, current_cost)
31            dfs(node.right, current_cost)
32
33    dfs(root, 0)
34    return increments
35
36# Example usage
37print(min_increments(7, [1,5,2,2,3,3,1]))  # Output: 6

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.