如何理解二叉树中的以下递归函数？

Question

The following is a Binary search function (a root has a left and a right child) which I don't quite comprehend. In the code it returns a list that is the longest path in the binary tree. However for the part: return_path_left = list_longest_path(node.left) return_path_right = list_longest_path(node.right) if len(return_path_left) > len(return_path_right):

how can you compare two recursive call? For example, if the tree is

1 / \\ 2 list_longest_path(node.right) will surely return [] . But how do you compare list_longest_path(2) with [] ?

Someone help will be great.

def list_longest_path(node):
    """
    List the data in a longest path of node.

    @param BinaryTree|None node: tree to list longest path of
    @rtype: list[object]

    >>> list_longest_path(None)
    []
    >>> list_longest_path(BinaryTree(5))
    [5]
    >>> b1 = BinaryTree(7)
    >>> b2 = BinaryTree(3, BinaryTree(2), None)
    >>> b3 = BinaryTree(5, b2, b1)
    >>> list_longest_path(b3)
    [5, 3, 2]
    """
    if node is None:
        return []
    else:
        return_path_left = list_longest_path(node.left)
        return_path_right = list_longest_path(node.right)
        if len(return_path_left) > len(return_path_right):
            return [node.data] + return_path_left
        else:
            return [node.data] + return_path_right

Answer 1

list_longest_path(node.right) will surely return []. But how do you compare list_longest_path(2) with []?

When a recursive call like list_longest_path(2) is encountered, it gets pushed onto the call stack. As the call stack is a stack [and is thus last in first out] the current stack frame is halted and list_longest_path(2) is evaluated.

list_longest_path(2) is evaluated like so:

As both left and right nodes are None, return_path_left = []; return_path_right = []; So list_longest_path(2) = [2] + [] = [2]

Then the list_longest_path(2) stackframe is popped off the stack and the program resumes execution in the previous stackframe. We now have a simple value for list_longest_path(2) = [2] We then finish up the execution of this function len([2]) > len([]) so list_longest_path(1) = [1] + [2] = [1,2]