230. Kth Smallest Element in a BST-优快云博客

本文介绍了一种方法，通过中序遍历来找到二叉搜索树中的第K小元素，同时讨论了如何优化算法以适应频繁修改的二叉搜索树。通过调整遍历策略和使用额外数据结构，提高了查询效率。

题目：

Given a binary search tree, write a function kthSmallest to 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 frequently? How would you optimize the kthSmallest routine?

Hint:

1. Try to utilize the property of a BST.
2. What if you could modify the BST node's structure?
3. The optimal runtime complexity is O(height of BST).

链接： http://leetcode.com/problems/kth-smallest-element-in-a-bst/

题解：

用了比较笨的方法 - inorder traversal，做了一下剪枝。

Time Complexity - O(k)， Space Complexity - O(k)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    List<Integer> nodeVal = new ArrayList<Integer>();
    
    public int kthSmallest(TreeNode root, int k) {
        if(root == null)
            return 0;
        inorder(root, k);
        return nodeVal.get(k - 1);
    }
    
    private void inorder(TreeNode root, int k) {
        if(nodeVal.size() >= k)
            return;
        if(root == null)
            return;
        inorder(root.left, k);
        nodeVal.add(root.val);
        inorder(root.right, k);
        
    }
}

Update:

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    private int count = 0;
    private int res = 0;
    public int kthSmallest(TreeNode root, int k) {
        if(root == null)
            return 0;
        inorder(root, k);
        return res;
    }
    
    private void inorder(TreeNode root, int k) {
        if(root == null)
            return;
        if(count >= k)
            return;
        inorder(root.left, k); 
        if(count >= k)
            return;
        res = root.val;
        count++;
        inorder(root.right, k);
        
    }
}

二刷:

一开始想到是用inorder traversal，可以建立一个list，当list.size() == k的时候我们跳出，最后返回list.get(k - 1)。后来想到我们可以继续优化，并不需要使用list来保存结果。

Java:

Inorder traversal:

Time Complexity - O(k)， Space Complexity - O(k)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        List<Integer> res = new ArrayList<>();
        inOrder(res, root, k);
        return res.size() >= k ? res.get(k - 1) : 0;
    }
    
    private void inOrder(List<Integer> res, TreeNode root, int k) {
        if (root == null || res.size() >= k)  return;
        inOrder(res, root.left, k);
        res.add(root.val);
        inOrder(res, root.right, k);
    }
}

优化:

不用建立List，直接使用count来记录

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int count = 0;
    private int res = 0;
    
    public int kthSmallest(TreeNode root, int k) {
        inOrder(root, k);
        return res;
    }
    
    private void inOrder(TreeNode root, int k) {
        if (root == null || count >= k)  return;
        inOrder(root.left, k);
        count++;
        if (count == k) res = root.val;
        inOrder(root.right, k);
    }
}

三刷:

依然是in-order traversal，写了iterative的版本。 Follow up在查询多和更改多的情况下，我们可以为BST的每个节点增加一个int count，然后进行二分查找。

Java:

Time Complexity - O(k)， Space Complexity - O(k)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        if (root == null) return Integer.MAX_VALUE;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        stack.push(node);
        while (node != null || !stack.isEmpty()) {
            if (node != null) {
                stack.push(node);
                node = node.left;
            } else {
                node = stack.pop();
                k--;
                if (k == 0) return node.val;
                node = node.right;
            }
        }
        return root.val;
    }
}

Reference:

https://leetcode.com/discuss/68052/two-easiest-in-order-traverse-java

https://leetcode.com/discuss/43267/4-lines-in-c

https://leetcode.com/discuss/43464/what-if-you-could-modify-the-bst-nodes-structure

https://leetcode.com/discuss/43299/o-k-space-o-n-time-10-short-lines-3-solutions

https://leetcode.com/discuss/45684/share-my-c-iterative-alg

https://leetcode.com/discuss/43771/implemented-java-binary-search-order-iterative-recursive

转载于:https://www.cnblogs.com/yrbbest/p/4999289.html