查找二叉树中的所有路径

Question

我正在尝试解决“给定二叉树，返回所有从根到叶的路径”的编码问题。

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

我见过一种解决方案

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        allPath = []
        if root is None:
            return []
        self.find_all_paths_recursive(root, [], allPath)
        return allPath

    def find_all_paths_recursive(self, currNode, currPath, allPath):
        if currNode is None:
            return 
        currPath.append(currNode.val)
        if currNode.left is None and currNode.right is None:
            currOut = '->'.join([str(x) for x in list(currPath)])
            allPath.append(currOut)
        # traverse left sub tree
        self.find_all_paths_recursive(currNode.left, currPath, allPath)
        # traverse right sub tree
        self.find_all_paths_recursive(currNode.right, currPath, allPath)
        del currPath[-1]

运行上面的代码给了我["1->2->5", "1->3"]的答案，这是正确的。

我正在考虑将if currNode.left is None and currNode.right is None:下的代码块更改为

      if currNode.left is None and currNode.right is None:
            #currOut = '->'.join([str(x) for x in list(currPath)])
            #allPath.append(currOut)
            allPath.append(currPath)

应该给我结果 [[1,2,5], [1,3]]。但是，此更改为我返回了结果 [[],[]]。我想知道为什么这不起作用？

Answer 1

由于您想直接添加 currPath，因此您必须在该时刻添加 currPath 的副本。

像这样：

if currNode.left is None and currNode.right is None:
    #currOut = '->'.join([str(x) for x in list(currPath)])
    # allPath.append(currOut)
    allPath.append(list(currPath))

编辑：

如果不添加 list，您会将原始列表对象添加到 allPath，该对象将由于递归而更新。添加list 将生成copy of the original list object，该copy of the original list object 将被保存且不会进一步更新。

Answer 2

递归是一种函数式遗产，因此将其与函数式风格一起使用会产生最佳效果。这意味着要避免变量重新分配、其他突变和副作用。

让我们看看以这种方式实现btree 会是什么样子。注意node 属性left 和right 可以在构造新节点时设置-

# btree.py

def empty():
  return None

class node:
  def __init__(self, val, left = empty(), right = empty()):
    self.val = val
    self.left = left
    self.right = right

def paths(t, p = ()):
  if not t:
    return
  elif t.left or t.right:
    yield from paths(t.left, (*p, t.val))
    yield from paths(t.right, (*p, t.val))
  else:
    yield "->".join(map(str, (*p, t.val)))

这是您现在的main 程序 -

# main.py

from btree import node, empty, paths

#    1
#  /   \
# 2     3
#  \
#   5

t = node(1, node(2, empty(), node(5)), node(3))

print(list(paths(t)))

['1->2->5', '1->3']