二叉树的每个叶子的路径

Question

上面的函数AllPaths（）将包含二叉树每个叶的路径的数组附加到全局数组res。

代码工作正常，但我想删除全局变量res，并使函数返回数组。我怎么能那样做？

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]]

Answer 1

您可以在递归中传递当前路径：

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

Answer 2

一种方法是通过回溯来实现：

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)

Answer 3

一个简单的方法可以让您完全避免内部列表和全局列表，就是创建一个生成器，在它们到来时生成值。然后，您可以将此传递到list以获得最终结果：

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]]