Path to each Leaf of a Binary Tree

user_10890520 0
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The function above AllPaths() appends an array containing the path to each leaf of the binary tree to the global array res.

The code works just fine, but I want to remove the global variable res and make the function return an array instead. How can I do that?

class Node:
    def __init__(self, value, left=None, right=None) -> None:
        self.value = value
        self.left  = left
        self.right = right

res = []
def allPaths(node, arr=[]):
    if node:
        tmp = [*arr, node.value]
        if not node.left and not node.right: # Leaf
            res.append(tmp)
        allPaths(node.left, tmp)
        allPaths(node.right, tmp)


root             = Node(1)
root.left        = Node(2);
root.left.left   = Node(4);
root.left.right  = Node(5);
root.right       = Node(3);
root.right.right = Node(6);
"""
          1         <-- root
        /   \
       2     3  
     /   \    \
    4     5    6    <-- leaves
"""
allPaths(root)
print(res)
# Output : [[1, 2, 4], [1, 2, 5], [1, 3, 6]]
binary-tree tree algorithm python
4 Answers

A simple way that allow you to avoid the inner lists and global list altogether is to make a generator that yields the values as they come. Then you can just pass this to list to make the final outcome:

class Node:
    def __init__(self, value, left=None, right=None) -> None:
        self.value = value
        self.left  = left
        self.right = right

def allPaths(node):  
    if node:
        if not node.left and not node.right: # Leaf
            yield [node.value]
        else:
            yield from ([node.value] + arr for arr in allPaths(node.left))
            yield from ([node.value] + arr for arr in allPaths(node.right))
              
root             = Node(1)
root.left        = Node(2);
root.left.left   = Node(4);
root.left.right  = Node(5);
root.right       = Node(3);
root.right.right = Node(6);
        
g = allPaths(root)
list(g)

# [[1, 2, 4], [1, 2, 5], [1, 3, 6]]
user_38746230

One method is to do it by backtracking:

def allPaths(node, partial_res, res):
    if not node: 
        return
    if not node.left and not node.right:
        res.append(partial_res[:] + [node.value])
        return    
    partial_res.append(node.value)
    allPaths(node.left, partial_res, res)
    allPaths(node.right, partial_res, res)
    partial_res.pop()

res = []
allPaths(root, [], res)
print(res)
user_77556650

You could pass down the current path in the recursion:

def allPaths(node,path=[]):
    if not node: return            # no node, do nothing
    fullPath = path + [node.value]
    if node.left or node.right:    # node is not a leaf, recurse down      
        yield from allPaths(node.left, fullPath)    # left leaves if any
        yield from allPaths(node.right, fullPath)   # right leaves if any
    else:
        yield fullPath             # leaf node, return final path
user_52375600

I offer another option.

def allPaths(root, path=[]):
    tmp = []
    if root.left:
        tmp.extend(allPaths(root.left, path + [root.value]))
    if root.right:
        tmp.extend(allPaths(root.right, path + [root.value]))
    if not root.left and not root.right:
        tmp.append(path + [root.value])
    return tmp


tree = allPaths(root)
print(tree)

The output is:

[[1, 2, 4], [1, 2, 5], [1, 3, 6]]
user_160201460
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