# Sales Path

The car manufacturer Honda holds their distribution system in the form of a tree (not necessarily binary). The root is the company itself, and every node in the tree represents a car distributor that receives cars from the parent node and ships them to its children nodes. The leaf nodes are car dealerships that sell cars direct to consumers. In addition, every node holds an integer that is the cost of shipping a car to it.

Take for example the tree below:

A path from Honda’s factory to a car dealership, which is a path from the root to a leaf in the tree, is called a Sales Path. The cost of a Sales Path is the sum of the costs for every node in the path. For example, in the tree above one Sales Path is 0→3→0→10, and its cost is `13` (0+3+0+10).

Honda wishes to find the minimal Sales Path cost in its distribution tree. Given a node `rootNode`, write a function `getCheapestCost`that calculates the minimal Sales Path cost in the tree.

Implement your function in the most efficient manner and analyze its time and space complexities.

For example:

Given the `rootNode` of the tree in diagram above

`7` since it’s the minimal Sales Path cost (there are actually two Sales Paths in the tree whose cost is `7`: 0→6→1 and 0→3→2→1→1)
`function getCheapestCost(rootNode):    n = rootNode.numberOfChildren()    if (n == 0):        return rootNode.cost    else:        # initialize minCost to the largest integer in the system        minCost = MAX_INT        for i from 0 to n-1:            tempCost = getCheapestCost(rootNode.child[i])            if (tempCost < minCost):                minCost = tempCost    return minCost + rootNode.cost`
Time Complexity: let `N` be the number of nodes in the tree. Notice that `getCheapestCost` is applied to every node exactly once. Therefore, there are overall `O(N)` calls to `getCheapestCost`.
Space Complexity: every time the function recurses, it consumes only a constant amount of space. However, due to the nature of the recursion we used, the stack call holds `N`instances of `getCheapestCost` which makes the total space complexity to be `O(N)`.