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子的路径的数组追加到全局数组 res。

代码工作正常，但我想删除全局变量 res 并将函数 return 改为数组。我该怎么做？

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

一种让您完全避免使用内部列表和全局列表的简单方法是制作一个生成器，在值出现时生成值。然后你可以将它传递给 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]]

Answer 2

一种方法是回溯：

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

你可以在递归中传递当前路径：

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

Answer 4

我提供另一种选择。

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)

输出为：

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

二叉树的每个叶子的路径

Path to each Leaf of a Binary Tree

python

algorithm

tree

binary-tree