Path Sum IV

Problem Statement

In this problem, you are given an array of three-digit integers known as nums, where each integer uniquely represents a node in a binary tree with a maximum depth of less than 5. The structure of each three-digit integer is as follows:

• The first digit (hundreds digit) indicates the depth d of the node within the tree, adhering to the constraint 1 <= d <= 4.
• The second digit (tens digit) specifies the position p of the node on its particular level. This is based on the node's position within a full binary tree, where the possible positions range from 1 to 8.
• The third digit (units digit) represents the actual value v of that node, ranging from 0 to 9.

You need to determine the sum of all the path values from the root of the tree to its leaves. Each path's value is calculated from the sequence of nodes' values that you traverse from the root to a leaf node. The array you are given is guaranteed to represent a valid binary tree and is sorted in ascending order based on the depth and position of the nodes.

Examples

Example 1

Input:

nums = [113,215,221]

Output:

12

Explanation:

The tree that the list represents is shown.

Example 2

Input:

nums = [113,221]

Output:

4

Explanation:

The tree that the list represents is shown.

Constraints

• 1 <= nums.length <= 15
• 110 <= nums[i] <= 489
• nums represents a valid binary tree with depth less than 5.
• nums is sorted in ascending order.

Approach and Intuition

1. Understanding the Tree Structure:

   • The information about node depth, position, and value are all encoded into a three-digit number which facilitates the approach to reconstructing the tree dynamically if needed.

2. Reconstruction of Tree:

   • Create a node structure that can maintain information regarding its value and its child nodes.
   • Parse through the nums array, decode each number into its depth, positional, and value components, and appropriately place it within a virtual binary tree structure.

3. Calculating Sum of All Path Values:

   • Begin at the root node and explore all possible paths to leaf nodes.
   • As each path from the root to a leaf is traversed, keep a running sum of node values until the leaf is reached, then add the final sum to a total sum count.
   • Use a recursive depth-first search (DFS) approach to explore each path, which is a common method for this type of tree traversal problem.

4. Return the Total Sum:

   • After all paths have been explored and their sums calculated, the final step is to return the accumulated sum of all path values from the root to the leaves.

Considering the constraints provided, such as the maximum possible number of nodes is 15 and the depth is less than 5, the approach described should be computationally feasible.

Solutions

class Solution {
public:
    int calculatePathSum(vector<int>& numbers) {
        unordered_map<int, int> nodeValues;
        for (int num : numbers) {
            int key = num / 10;
            int value = num % 10;
            nodeValues[key] = value;
        }
    
        queue<pair<int, int>> nodesQueue;
        int sumResult = 0;
    
        int baseNode = numbers[0] / 10;
        nodesQueue.push({baseNode, nodeValues[baseNode]});
    
        while (!nodesQueue.empty()) {
            auto [nodeKey, accumulatedSum] = nodesQueue.front();
            nodesQueue.pop();
    
            int level = nodeKey / 10;
            int pos = nodeKey % 10;
    
            int leftChildKey = (level + 1) * 10 + pos * 2 - 1;
            int rightChildKey = (level + 1) * 10 + pos * 2;
    
            if (!nodeValues.count(leftChildKey) && !nodeValues.count(rightChildKey)) {
                sumResult += accumulatedSum;
            }
    
            if (nodeValues.count(leftChildKey)) {
                nodesQueue.push({leftChildKey, accumulatedSum + nodeValues[leftChildKey]});
            }
    
            if (nodeValues.count(rightChildKey)) {
                nodesQueue.push({rightChildKey, accumulatedSum + nodeValues[rightChildKey]});
            }
        }
    
        return sumResult;
    }
};