递归 function 不返回 python 中的数组

Question

I'm writing a simple code that returns the path to the destination node in BST.

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

root = TreeNode(6)
root.left = TreeNode(2)
root.right = TreeNode(8)
root.left.left = TreeNode(0)
root.left.right = TreeNode(4)
root.left.right.left = TreeNode(3)
root.left.right.right = TreeNode(5)
root.right.left = TreeNode(7)
root.right.right = TreeNode(9)

After defining the tree;

p = 2
q = 8

def pathFind(path, cur, node): # path : path the function walked throug so far
                               # cur : current node
                               # node : destination node's value 
  
  #print(f'current node value is {cur.val}')
  #print(path)
  ## ending condition ##
  if cur.val == node: # If we reach the destination node
    return path
  
  elif cur.val < node and cur.right != None : 
    # 'if cur.right != None:' line is useless since the problem guarantees the existence of destination value in BST 
    path.append(cur)

    return pathFind(path, cur.right, node)
  elif cur.val > node and cur.left != None: # cur.val > node:
    path.append(cur)
    return pathFind(path, cur.left, node)
  else:
    return None

path_p = pathFind([root], root, p) 

I checked that my function reaches the destination and record the path toward it without any problem, but the last line - path_p = pathFind([root], root, p) doesn't work.

Anyone could help?

Answer 1

In function pathFind() , the path is returned only by the execution where the tested node contains the target value. The (all) previous executions discard that return value. Fix it by putting return before the recursive calls.

Answer 2

Try below code to find the path for a target node

def pathFind(path, node, target):
    if node == None:
        return
    else:
        path.append(node.val)
        if node.val == target:
            return path
        elif node.val < target and node.right != None:
            return pathFind(path, node.right, target)
        elif node.val > target and node.left != None:
            return pathFind(path, node.left, target)
        else:
            return None

print(pathFind([], root, 9))

output

[6, 8, 9]

Answer 3

A few remaining issues (after the edits to your question):

• The root node is in most cases added twice to the path: once in the initial call, and again when moving to the left or right subtree of the root node: cur is the root node and path.append(cur) is executed.

  So, pass an empty list in the initial call instead of [root]

• When the target node is found, that node is not appended to the path, yet I suppose it should be the final node in the path. So path.append(cur) should also happen in the first if block.

Those changes will fix your code.