【leetcode】230. Kth Smallest Element in a BST 题目如下: 解题思路:本题对运行时间的要求比较低,我试过把遍历所有节点并且把节点的值存入数组,最后排序并取第K-1位的元素作为结果返回也能通过。我的方法是省去了最后排序的步骤,因为BST的规律就是 node.left.val < node.val < node.right.val,所以只要按...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public:intkthSmallest(TreeNode* root,intk) {intindex =0,num = -1; kthSmallest(root,k,index,num);returnnum; }voidkthSmallest(TreeNode* root,intk,int& index,int&num){if(!root)return; kthSmallest(ro...
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine? Hint: Try to utilize the property of a BST. What if you could modify the BST node's structure? The optimal runtime complexity is...
Given a binary search tree, write a functionkthSmallestto find the kth smallest element in it. Note: You may assume k is always valid, 1 ≤ k ≤ BST's total elements. Follow up: What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequ...
Leetcode 230. Kth Smallest Element in a BST 简介:先来看下二叉搜索树的性质,对于任意一个非叶子节点,它的左子树中所有的值必定小于这个节点的val,右子树中所有的值必定大于或等于当前节点的val。 这条性质就保证了如果我们对二叉搜索树做中序遍历,中序遍历的结果肯定是有序的。对于此题而言,我们只需要拿到...
题目地址:Kth Smallest Element in a BST 题目简介: 给定一个二叉搜索树,编写一个函数 kthSmallest 来查找其中第 k 个最小的元素。 说明:你可以假设 k 总是有效的,1 ≤ k ≤ 二叉搜索树元素个数。 示例 1: 示例 2: &n...猜你喜欢LeetCode——230. 二叉搜索树中第K小的元素(Kth Smallest Element in...
遍历数组时将数字加入优先队列(堆),一旦堆的大小大于k就将堆顶元素去除,确保堆的大小为k。遍历完后堆顶就是返回值。 代码 public class Solution { public int findKthLargest(int[] nums, int k) { PriorityQueue<Integer> p = new PriorityQueue<Integer>(); ...
Given the root of a binary search tree, and an integer k, returnthekth (1-indexed)smallest element in the tree. class Solution { public int kthSmallest(TreeNode root, int k) { Deque<TreeNode> stack = new LinkedList<>(); TreeNode cur = root; ...
Given a binary search tree, write a functionkthSmallestto find the kth smallest element in it. Note: You may assume k is always valid, 1 ≤ k ≤ BST's total elements. 问题:找出二叉搜索树种第 k 小的元素。 一个深度遍历的应用。使用递归、或者借助栈都可以实现深度遍历。本文代码使用递归实现...
下面的伪代码摘录自 http://bookshadow.com/weblog/2015/07/02/leetcode-kth-smallest-element-bst/ 假设BST节点TreeNode的属性能够扩展,则再加入一个属性leftCnt,记录左子树的节点个数 记当前节点为node 当node不为空时循环: 若k == node.leftCnt + 1:则返回node ...