如何使用Python实现链表的二叉搜索树迭代器操作
链表的二叉搜索树迭代器操作可以通过先将链表转换为二叉搜索树，然后再对二叉搜索树进行中序遍历（左节点-根节点-右节点）来实现。具体实现方法如下：
- 定义一个二叉搜索树节点类，包含节点值，左右子节点指针和父节点指针（为了方便遍历回退，需要记录父节点指针）。
class BSTNode: def __init__(self, value): self.value = value self.left = None self.right = None self.parent = None
- 定义一个转换链表为二叉搜索树的函数。由于链表是单向的，所以不能直接进行平衡二叉搜索树的构建，需要采用递归的方法，每次选择链表的中间节点作为二叉搜索树的根节点，然后递归构建左子树和右子树。为了方便递归操作，我们可以先定义一个转换数组为二叉搜索树的函数。
def array_to_bst(nums, start, end, parent): if start > end: return None mid = (start + end) // 2 node = BSTNode(nums[mid]) node.parent = parent node.left = array_to_bst(nums, start, mid - 1, node) node.right = array_to_bst(nums, mid + 1, end, node) return node def linkedlist_to_bst(head, tail, parent): if head == tail: return None slow, fast = head, head while fast != tail and fast.next != tail: slow = slow.next fast = fast.next.next node = BSTNode(slow.val) node.parent = parent node.left = linkedlist_to_bst(head, slow, node) node.right = linkedlist_to_bst(slow.next, tail, node) return node
- 定义一个二叉搜索树中序遍历迭代器的类。在初始化的时候，通过转换链表为二叉搜索树的函数创建一个二叉搜索树和一个栈。然后将二叉搜索树根节点及其所有左子节点入栈。每次取栈顶元素时，如果该节点的右子节点不为空，则将该节点的右子节点及其所有左子节点入栈，否则将栈顶元素的父节点及其所有左子节点入栈。
class BSTIterator: def __init__(self, head): self.root = linkedlist_to_bst(head, None, None) self.stack = [] node = self.root while node: self.stack.append(node) node = node.left def has_next(self): return len(self.stack) > 0 def next(self): node = self.stack.pop() val = node.value node = node.right while node: self.stack.append(node) node = node.left return val
- 演示代码：
class ListNode: def __init__(self, val=0, next=None): self.val = val self.next = next class BSTNode: def __init__(self, value): self.value = value self.left = None self.right = None self.parent = None def array_to_bst(nums, start, end, parent): if start > end: return None mid = (start + end) // 2 node = BSTNode(nums[mid]) node.parent = parent node.left = array_to_bst(nums, start, mid - 1, node) node.right = array_to_bst(nums, mid + 1, end, node) return node def linkedlist_to_bst(head, tail, parent): if head == tail: return None slow, fast = head, head while fast != tail and fast.next != tail: slow = slow.next fast = fast.next.next node = BSTNode(slow.val) node.parent = parent node.left = linkedlist_to_bst(head, slow, node) node.right = linkedlist_to_bst(slow.next, tail, node) return node class BSTIterator: def __init__(self, head): self.root = linkedlist_to_bst(head, None, None) self.stack = [] node = self.root while node: self.stack.append(node) node = node.left def has_next(self): return len(self.stack) > 0 def next(self): node = self.stack.pop() val = node.value node = node.right while node: self.stack.append(node) node = node.left return val if __name__ == "__main__": linkedlist = ListNode('p', ListNode('i', ListNode('d', ListNode('a', ListNode('n', ListNode('c', ListNode('o', ListNode('d', ListNode('e', ListNode('.', None)))))))))) bst_iterator = BSTIterator(linkedlist) while bst_iterator.has_next(): print(bst_iterator.next())
其中，链表转换为二叉搜索树的函数中用到了快慢指针求链表中间节点的方法。该方法可以起到快速寻找链表中间节点的作用，避免了遍历链表的时间复杂度为O(n)。
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