zoukankan      html  css  js  c++  java
  • #leetcode#Kth Smallest Element in a BST

    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?


    思路, in order traversal,

    注意第一个元素的推断, 设root为当前节点, 则 root.left == null && count == 0,就保证了root为 1st smallest element。样例:


       4

      /

    /

         2

      /

    /

         1



        4

      /

    /

           2

             

               

        3


    上面两个bst中第一个节点分别为 1 和 2


    /**
     * 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> count = new ArrayList<Integer>();
            count.add(0);
            helper(root, count, k);
            return count.get(1);
        }
        
        private void helper(TreeNode root, List<Integer> count, int target){
            if(root == null){
                return;
            }
            
            helper(root.left, count, target);
            
            if(root.left == null && count.get(0) == 0){
                count.set(0, 1);
            }else{
                count.set(0, count.get(0) + 1);
            }
            if(count.get(0) == target){
                count.add(root.val);
                return;
            }
            
            helper(root.right, count, target);
        }
    }


  • 相关阅读:
    枚举
    泛型
    装箱和拆箱
    使用TryParse()来执行数值转换
    参数数组
    checked和unchecked转换
    字符串不可变
    TCC : Tiny C Compiler (2018-2-6)
    win10 下 protobuf 与 qt
    QWebView 与Js 交互
  • 原文地址:https://www.cnblogs.com/zfyouxi/p/5224331.html
Copyright © 2011-2022 走看看