Python 存储在一维数组中的树代码包含与右节点指针相关的逻辑错误

Question

在我正在使用的 CS 教科书的以下代码中，在插入值32后的树的RightPointer中，数组 position 0中节点的 RightPointer 设置为3 。 这对我来说没有意义，因为我认为它仍然应该是2 。

我看起来代码不正确，因为现在有两个RightPointer为3的节点。 谁能解释一下我需要做什么才能使代码正常工作？

我猜这条线是罪魁祸首： tree[PreviousNodePtr].RightPointer = NewNodePtr因为那是RightPointer更新的地方，但我看不出我会如何改变它来避免这个问题。

# NULL should be set to -1 if using array element with index 0
NULL = -1


# Declare record type to store data and pointer
class TreeNode:
    def __init__(self):
        self.Data = ""
        self.LeftPointer = NULL
        self.RightPointer = NULL


def print_tree(tree):
    print("{0:^5}|{1:^10}|{2:^10}|{3:^10}|".format("", "Left", "Data", "Right"))
    for i in range(len(tree)):
        index = "[" + str(i) + "]"
        print("{0:^5}|{1:^10}|{2:^10}|{3:^10}|".format(index, tree[i].LeftPointer, tree[i].Data, tree[i].RightPointer))

    print("RootPointer", RootPointer, "FreePtr", FreePtr, "\n")


def InsertNode(tree, NewItem):
    global RootPointer, FreePtr

    if FreePtr == NULL:
        # tree is full
        print("Tree is full")
        return

    # There is space in the array
    # Take node from free list and store data item
    NewNodePtr = FreePtr
    tree[NewNodePtr].Data = NewItem
    FreePtr = tree[FreePtr].LeftPointer
    tree[NewNodePtr].LeftPointer = NULL
    # check if empty tree
    if RootPointer == NULL:
        # insert new node at root
        RootPointer = NewNodePtr
    else:  # find insertion point
        ThisNodePtr = RootPointer
        while ThisNodePtr != NULL:  # while not a leaf node
            PreviousNodePtr = ThisNodePtr  # remember this node
            if tree[ThisNodePtr].Data > NewItem:
                TurnedLeft = True  # # follow left pointer
                ThisNodePtr = tree[ThisNodePtr].LeftPointer
            else:
                TurnedLeft = False
                ThisNodePtr = tree[ThisNodePtr].RightPointer
            if TurnedLeft:
                tree[PreviousNodePtr].LeftPointer = NewNodePtr
            else:
                tree[PreviousNodePtr].RightPointer = NewNodePtr


tree = [TreeNode() for i in range(5)]
RootPointer = NULL  # set Root pointer
FreePtr = 0  # set starting position of list
for i in range(len(tree) - 1):  # link all nodes to make free list
    tree[i].LeftPointer = i + 1

print_tree(tree)
InsertNode(tree, 10)
print_tree(tree)
InsertNode(tree, 5)
print_tree(tree)
InsertNode(tree, 20)
print_tree(tree)
InsertNode(tree, 32)
print_tree(tree)

Output：

     |   Left   |   Data   |  Right   |
 [0] |    1     |          |    -1    |
 [1] |    2     |          |    -1    |
 [2] |    3     |          |    -1    |
 [3] |    4     |          |    -1    |
 [4] |    -1    |          |    -1    |
RootPointer -1 FreePtr 0

     |   Left   |   Data   |  Right   |
 [0] |    -1    |    10    |    -1    |
 [1] |    2     |          |    -1    |
 [2] |    3     |          |    -1    |
 [3] |    4     |          |    -1    |
 [4] |    -1    |          |    -1    |
RootPointer 0 FreePtr 1

     |   Left   |   Data   |  Right   |
 [0] |    1     |    10    |    -1    |
 [1] |    -1    |    5     |    -1    |
 [2] |    3     |          |    -1    |
 [3] |    4     |          |    -1    |
 [4] |    -1    |          |    -1    |
RootPointer 0 FreePtr 2

     |   Left   |   Data   |  Right   |
 [0] |    1     |    10    |    2     |
 [1] |    -1    |    5     |    -1    |
 [2] |    -1    |    20    |    -1    |
 [3] |    4     |          |    -1    |
 [4] |    -1    |          |    -1    |
RootPointer 0 FreePtr 3

     |   Left   |   Data   |  Right   |
 [0] |    1     |    10    |    3     |
 [1] |    -1    |    5     |    -1    |
 [2] |    -1    |    20    |    3     |
 [3] |    -1    |    32    |    -1    |
 [4] |    -1    |          |    -1    |
RootPointer 0 FreePtr 4

Answer 1

问题是您没有更新您的ThisNodePtr 。 所以ThisNodePtr指针不会向左或向右移动，因此它总是更新根的左或右。

if TurnedLeft:
    if tree[PreviousNodePtr].LeftPointer != NULL:
        PreviousNodePtr = ThisNodePtr
        ThisNodePtr = tree[ThisNodePtr].LeftPointer
        tree[PreviousNodePtr].LeftPointer = NewNodePtr
    else:
        tree[PreviousNodePtr].LeftPointer = NewNodePtr 
else:
    if tree[PreviousNodePtr].RightPointer != NULL:
        PreviousNodePtr = ThisNodePtr
        ThisNodePtr = tree[ThisNodePtr].RightPointer
        tree[PreviousNodePtr].RightPointer = NewNodePtr
    else:
        tree[PreviousNodePtr].RightPointer = NewNodePtr

如果需要，您也可以使用InsertNode(tree, 3)而不是InsertNode(tree, 32)对其进行测试。 当左节点不是NULL时，它将向左移动。