构建一颗二叉查找树并实现其前序、中序、后序及层序遍

二叉查找树是一种特殊的二叉树，其中每个节点的值比其左子树中的值大，比其右子树中的值小。下面是构建一颗二叉查找树的Python代码：

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

def insert(root, val):
    if root is None:
        return TreeNode(val)
    elif val < root.val:
        root.left = insert(root.left, val)
    else:
        root.right = insert(root.right, val)
    return root

def preorder(root):
    if root is None:
        return
    print(root.val, end=' ')
    preorder(root.left)
    preorder(root.right)

def inorder(root):
    if root is None:
        return
    inorder(root.left)
    print(root.val, end=' ')
    inorder(root.right)

def postorder(root):
    if root is None:
        return
    postorder(root.left)
    postorder(root.right)
    print(root.val, end=' ')

def levelorder(root):
    if root is None:
        return
    queue = [root]
    while queue:
        node = queue.pop(0)
        print(node.val, end=' ')
        if node.left:
            queue.append(node.left)
        if node.right:
            queue.append(node.right)

# 构建二叉查找树
root = None
values = [8, 3, 10, 1, 6, 14, 4, 7, 13]
for val in values:
    root = insert(root, val)

print("前序遍历结果：")
preorder(root)
print("\n中序遍历结果：")
inorder(root)
print("\n后序遍历结果：")
postorder(root)
print("\n层序遍历结果：")
levelorder(root)

运行结果如下：

前序遍历结果：
8 3 1 6 4 7 10 14 13 
中序遍历结果：
1 3 4 6 7 8 10 13 14 
后序遍历结果：
1 4 7 6 3 13 14 10 8 
层序遍历结果：
8 3 10 1 6 14 4 7 13 
``