LeetCode每日一题(173. Binary Search Tree Iterator)

Implement the BSTIterator class that represents an iterator over the in-order traversal of a binary search tree (BST):

BSTIterator(TreeNode root) Initializes an object of the BSTIterator class. The root of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
boolean hasNext() Returns true if there exists a number in the traversal to the right of the pointer, otherwise returns false.
int next() Moves the pointer to the right, then returns the number at the pointer.
Notice that by initializing the pointer to a non-existent smallest number, the first call to next() will return the smallest element in the BST.

You may assume that next() calls will always be valid. That is, there will be at least a next number in the in-order traversal when next() is called.

Example 1:

Input
[“BSTIterator”, “next”, “next”, “hasNext”, “next”, “hasNext”, “next”, “hasNext”, “next”, “hasNext”]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
Output
[null, 3, 7, true, 9, true, 15, true, 20, false]

Explanation

BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // return 3
bSTIterator.next(); // return 7
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 9
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 15
bSTIterator.hasNext(); // return True
bSTIterator.next(); // return 20
bSTIterator.hasNext(); // return False

Constraints:

• The number of nodes in the tree is in the range [1, 105].
• 0 <= Node.val <= 106
• At most 105 calls will be made to hasNext, and next.

---

In-Order 就是先左后自己最后右, 对于 BST 来说最后的输出结果就是从小到大， 说到这，首先想到的最简单的办法就是遍历整棵树然后收集成一个 BinaryHeap, 后面就是不停的 pop 就好了。但是既然是处理树的问题， 用递归的思想貌似看上去更地道一点。

---

use std::cell::RefCell;
use std::rc::Rc;

struct BSTIterator {
    root: Option<Rc<RefCell<TreeNode>>>,
}

fn next(root: Option<Rc<RefCell<TreeNode>>>) -> (i32, Option<Rc<RefCell<TreeNode>>>) {
    if let Some(node) = &root {
        let (left_val, left) = next(node.borrow_mut().left.take());
        node.borrow_mut().left = left;
        if left_val >= 0 {
            return (left_val, Some(node.clone()));
        }
        let right = node.borrow_mut().right.take();
        return (node.borrow().val, right);
    }
    (-1, None)
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl BSTIterator {
    fn new(root: Option<Rc<RefCell<TreeNode>>>) -> Self {
        Self { root }
    }

    fn next(&mut self) -> i32 {
        let (val, root) = next(self.root.take());
        self.root = root;
        val
    }

    fn has_next(&self) -> bool {
        self.root.is_some()
    }
}