190 lines
3.6 KiB
Markdown
190 lines
3.6 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/kth-smallest-element-in-a-bst/description/
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## 题目描述
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```
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Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.
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Note:
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You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
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Example 1:
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Input: root = [3,1,4,null,2], k = 1
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3
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/ \
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1 4
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\
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2
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Output: 1
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Example 2:
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Input: root = [5,3,6,2,4,null,null,1], k = 3
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5
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/ \
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3 6
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/ \
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2 4
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/
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1
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Output: 3
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Follow up:
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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?
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```
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## 思路
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解法一：
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由于‘中序遍历一个二叉查找树（BST）的结果是一个有序数组’ ，因此我们只需要在遍历到第k个，返回当前元素即可。
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中序遍历相关思路请查看[binary-tree-traversal](../thinkings/binary-tree-traversal.md)
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解法二：
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联想到二叉搜索树的性质，root 大于左子树，小于右子树，如果左子树的节点数目等于 K-1，那么 root 就是结果，否则如果左子树节点数目小于 K-1，那么结果必然在右子树，否则就在左子树。
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因此在搜索的时候同时返回节点数目，跟 K 做对比，就能得出结果了。
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## 关键点解析
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- 中序遍历
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## 代码
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解法一：
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JavaScript Code:
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```js
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/*
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* @lc app=leetcode id=230 lang=javascript
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*
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* [230] Kth Smallest Element in a BST
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*/
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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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/**
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* @param {TreeNode} root
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* @param {number} k
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* @return {number}
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*/
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var kthSmallest = function(root, k) {
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const stack = [root];
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let cur = root;
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let i = 0;
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function insertAllLefts(cur) {
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while(cur && cur.left) {
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const l = cur.left;
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stack.push(l);
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cur = l;
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}
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}
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insertAllLefts(cur);
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while(cur = stack.pop()) {
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i++;
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if (i === k) return cur.val;
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const r = cur.right;
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if (r) {
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stack.push(r);
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insertAllLefts(r);
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}
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}
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return -1;
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};
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```
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Java Code:
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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private int count = 1;
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private int res;
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public int KthSmallest (TreeNode root, int k) {
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inorder(root, k);
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return res;
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}
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public void inorder (TreeNode root, int k) {
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if (root == null) return;
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inorder(root.left, k);
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if (count++ == k) {
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res = root.val;
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return;
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}
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inorder(root.right, k);
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}
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```
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解法二：
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JavaScript Code:
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```js
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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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function nodeCount(node) {
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if (node === null) return 0;
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const l = nodeCount(node.left);
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const r = nodeCount(node.right);
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return 1 + l + r;
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}
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/**
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* @param {TreeNode} root
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* @param {number} k
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* @return {number}
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*/
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var kthSmallest = function(root, k) {
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const c = nodeCount(root.left);
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if (c === k - 1) return root.val;
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else if (c < k - 1) return kthSmallest(root.right, k - c - 1);
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return kthSmallest(root.left, k)
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};
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```
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## 扩展
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这道题有一个follow up：
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`What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently?
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How would you optimize the kthSmallest routine?`
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大家可以思考一下。
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