How do I get a multiple of leaf nodes for each element in dict as tree structure?

Question

Let’s say here is input:

input = {1:[2,3], 2:[], 3:[4,5,6], 4:[], 5:[], 6:[]}

This can be represented like below:
tree

If all leaf nodes are 1, we can change like this:

for idx, val in input[1]:
    if len(input[val]) == 0:
        input[1][idx] = [] # input[1] = [[], 3]
    else:
        input[1][idx] = input[val] # input[1] = [[], [4,5,6]]

And then somehow, input[1] can be input[1] = [[],[[],[],[]]].

So finally, i wanna get multiple numbers of each dict key elements compared to leaf nodes.

I am not sure the description that i wrote is clear.

Anyways, What i want to get is like:

# 2, 4, 5, 6 are all leaf nodes.
# 3 is including [4, 5, 6] which are all leaf nodes, then 3's value must be 3 (len[4,5,6])
# 1 is including [2, 3]. Among them, only 2 is leaf. And 3's value is 3. So output[1] = 1 + 3 = 4

output = {1:4, 2:1, 3:3, 4:1, 5:1, 6:1}

Asked By: user20149880

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Source

Answer 1

Simple recursion with functools.lru_cache:

from functools import lru_cache


def leaves_count(tree):
    @lru_cache
    def cntr(key):
        value = tree[key]
        return sum(map(cntr, value)) if value else 1

    return {k: cntr(k) for k in tree}

Test:

>>> tree = {1: [2, 3], 2: [], 3: [4, 5, 6], 4: [], 5: [], 6: []}
>>> leaves_count(tree)
{1: 4, 2: 1, 3: 3, 4: 1, 5: 1, 6: 1}

Manually implement the cached version:

def leaves_count(tree):
    def cntr(key):
        cnt = cache.get(key)
        if cnt:
            return cnt

        value = tree[key]
        cache[key] = cnt = sum(map(cntr, value)) if value else 1
        return cnt

    cache = {}
    return {k: cntr(k) for k in tree}

Answered By: Mechanic Pig

Answer 2

Using recursion method

Code:

dic=   {1:[2,3], 2:[], 3:[4,5,6], 4:[], 5:[], 6:[]}

def recur(val, leaf):
    for v in val:
        if v in dic.keys():
            if len(dic[v])==0:
                leaf.append(v)
            else:
                recur(dic[v],leaf) 
        else:
            leaf.append(v)
    return len(leaf)
            

{key : 1 if recur(val,[])==0 else recur(val,[]) for key,val in dic.items()}

Output:

{1: 4, 2: 1, 3: 3, 4: 1, 5: 1, 6: 1}

Input: {1:[2,3], 2:[], 3:[4,5,6], 4:[], 5:[], 6:[7,8], 7:[], 8:[9,10], 9:[]}

Output: {1: 6, 2: 1, 3: 5, 4: 1, 5: 1, 6: 3, 7: 1, 8: 2, 9: 1}

Answered By: Stackpy

How do I get a multiple of leaf nodes for each element in dict as tree structure?

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