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Camino a cada hoja de un árbol binario

La función anterior AllPaths() agrega una matriz que contiene la ruta a cada hoja del árbol binario a la matriz global res .

El código funciona bien, pero quiero eliminar la variable global res y hacer que la función devuelva una matriz en su lugar. ¿Cómo puedo hacer eso?

 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]]
over 4 years ago · Santiago Trujillo
4 answers
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0

Una forma simple que le permite evitar las listas internas y la lista global por completo es hacer un generador que produzca los valores tal como vienen. Luego, puede pasar esto a la list para obtener el resultado final:

 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]]
over 4 years ago · Santiago Trujillo Report

0

Un método es hacerlo retrocediendo:

 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)
over 4 years ago · Santiago Trujillo Report

0

Podría pasar la ruta actual en la recursividad:

 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
over 4 years ago · Santiago Trujillo Report

0

Ofrezco otra opción.

 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)

La salida es:

 [[1, 2, 4], [1, 2, 5], [1, 3, 6]]
over 4 years ago · Santiago Trujillo Report
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