173. 二叉搜索树迭代器

题目描述

实现一个二叉搜索树迭代器类BSTIterator ，表示一个按中序遍历二叉搜索树（BST）的迭代器：

BSTIterator(TreeNode root) 初始化 BSTIterator 类的一个对象。BST 的根节点 root 会作为构造函数的一部分给出。指针应初始化为一个不存在于 BST 中的数字，且该数字小于 BST 中的任何元素。
boolean hasNext() 如果向指针右侧遍历存在数字，则返回 true ；否则返回 false 。
int next()将指针向右移动，然后返回指针处的数字。

注意，指针初始化为一个不存在于 BST 中的数字，所以对 next() 的首次调用将返回 BST 中的最小元素。

你可以假设 next() 调用总是有效的，也就是说，当调用 next() 时，BST 的中序遍历中至少存在一个下一个数字。

示例：

输入
["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
[[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
输出
[null, 3, 7, true, 9, true, 15, true, 20, false]
解释
BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
bSTIterator.next(); // 返回 3
bSTIterator.next(); // 返回 7
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 9
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 15
bSTIterator.hasNext(); // 返回 True
bSTIterator.next(); // 返回 20
bSTIterator.hasNext(); // 返回 False

提示：

树中节点的数目在范围 [1, 10⁵] 内
0 <= Node.val <= 10⁶
最多调用 10⁵ 次 hasNext 和 next 操作

进阶：

你可以设计一个满足下述条件的解决方案吗？next() 和 hasNext() 操作均摊时间复杂度为 O(1) ，并使用 O(h) 内存。其中 h 是树的高度。

解法

方法一：递归

初始化数据时，递归中序遍历，将二叉搜索树每个结点的值保存在列表 vals 中。用 cur 指针记录外部即将遍历的位置，初始化为 0。

调用 next() 时，返回 vals[cur]，同时 cur 指针自增。调用 hasNext() 时，判断 cur 指针是否已经达到 len(vals) 个数，若是，说明已经遍历结束，返回 false，否则返回 true。

方法二：栈迭代

初始化时，从根节点一路遍历所有左子节点，压入栈 stack 中。

调用 next()时，弹出栈顶元素 cur，获取 cur 的右子节点 node，若 node 不为空，一直循环压入左节点。最后返回 cur.val 即可。调用 hasNext() 时，判断 stack 是否为空，空则表示迭代结束。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class BSTIterator:

    def __init__(self, root: TreeNode):
        def inorder(root):
            if root:
                inorder(root.left)
                self.vals.append(root.val)
                inorder(root.right)

        self.cur = 0
        self.vals = []
        inorder(root)

    def next(self) -> int:
        res = self.vals[self.cur]
        self.cur += 1
        return res

    def hasNext(self) -> bool:
        return self.cur < len(self.vals)


# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class BSTIterator:

    def __init__(self, root: TreeNode):
        self.stack = []
        while root:
            self.stack.append(root)
            root = root.left

    def next(self) -> int:
        cur = self.stack.pop()
        node = cur.right
        while node:
            self.stack.append(node)
            node = node.left
        return cur.val

    def hasNext(self) -> bool:
        return len(self.stack) > 0


# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class BSTIterator {
    private int cur = 0;
    private List<Integer> vals = new ArrayList<>();

    public BSTIterator(TreeNode root) {
        inorder(root);
    }

    public int next() {
        return vals.get(cur++);
    }

    public boolean hasNext() {
        return cur < vals.size();
    }

    private void inorder(TreeNode root) {
        if (root != null) {
            inorder(root.left);
            vals.add(root.val);
            inorder(root.right);
        }
    }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class BSTIterator {
    private Deque<TreeNode> stack = new LinkedList<>();

    public BSTIterator(TreeNode root) {
        for (; root != null; root = root.left) {
            stack.offerLast(root);
        }
    }

    public int next() {
        TreeNode cur = stack.pollLast();
        for (TreeNode node = cur.right; node != null; node = node.left) {
            stack.offerLast(node);
        }
        return cur.val;
    }

    public boolean hasNext() {
        return !stack.isEmpty();
    }
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator obj = new BSTIterator(root);
 * int param_1 = obj.next();
 * boolean param_2 = obj.hasNext();
 */

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BSTIterator {
public:
    vector<int> vals;
    int cur;
    BSTIterator(TreeNode* root) {
        cur = 0;
        inorder(root);
    }

    int next() {
        return vals[cur++];
    }

    bool hasNext() {
        return cur < vals.size();
    }

    void inorder(TreeNode* root) {
        if (root) {
            inorder(root->left);
            vals.push_back(root->val);
            inorder(root->right);
        }
    }
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BSTIterator {
public:
    stack<TreeNode*> stack;
    BSTIterator(TreeNode* root) {
        for (; root != nullptr; root = root->left) {
            stack.push(root);
        }
    }

    int next() {
        TreeNode* cur = stack.top();
        stack.pop();
        TreeNode* node = cur->right;
        for (; node != nullptr; node = node->left) {
            stack.push(node);
        }
        return cur->val;
    }

    bool hasNext() {
        return !stack.empty();
    }
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
type BSTIterator struct {
	stack []*TreeNode
}

func Constructor(root *TreeNode) BSTIterator {
	var stack []*TreeNode
	for ; root != nil; root = root.Left {
		stack = append(stack, root)
	}
	return BSTIterator{
		stack: stack,
	}
}

func (this *BSTIterator) Next() int {
	cur := this.stack[len(this.stack)-1]
	this.stack = this.stack[:len(this.stack)-1]
	for node := cur.Right; node != nil; node = node.Left {
		this.stack = append(this.stack, node)
	}
	return cur.Val
}

func (this *BSTIterator) HasNext() bool {
	return len(this.stack) > 0
}

/**
 * Your BSTIterator object will be instantiated and called as such:
 * obj := Constructor(root);
 * param_1 := obj.Next();
 * param_2 := obj.HasNext();
 */

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 */
var BSTIterator = function (root) {
    this.stack = [];
    for (; root != null; root = root.left) {
        this.stack.push(root);
    }
};

/**
 * @return {number}
 */
BSTIterator.prototype.next = function () {
    let cur = this.stack.pop();
    let node = cur.right;
    for (; node != null; node = node.left) {
        this.stack.push(node);
    }
    return cur.val;
};

/**
 * @return {boolean}
 */
BSTIterator.prototype.hasNext = function () {
    return this.stack.length > 0;
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * var obj = new BSTIterator(root)
 * var param_1 = obj.next()
 * var param_2 = obj.hasNext()
 */

buelev / leetcode

173. 二叉搜索树迭代器

题目描述

解法

Python3

Java

C++

Go

JavaScript

...

简介

发行版

贡献者

近期动态

buelev / leetcode .gitee-modal { width: 500px !important; }

173. 二叉搜索树迭代器

题目描述

解法

Python3

Java

C++

Go

JavaScript

...

简介

发行版

贡献者

近期动态

搜索帮助

buelev / leetcode