Kth smallest element (in a sorted matrix ; In a BST tree ; In two sorted array)

包含了三个问题

1.Kth smallest element in a sorted matrix

2.Kth smallest element in a BST(Binary Search Tree)

3.Kth smallest element in two sorted array

/** * 包含了三个问题 * 1.Kth smallest element in a sorted matrix * 2.Kth smallest element in a BST(Binary Search Tree) * 3.Kth smallest element in two sorted array * @author 健身小码哥 * */public class KthSmallestElement {/** * Kth smallest element in a sorted matrix * 思路一：二分查找 * @param matrix * @param k * @return */public int findKthSmallestElementInSortedMatrix(int matrix[][] , int k){int n = matrix.length;        int startValue = matrix[0][0];        int endValue = matrix[n-1][n-1];                while (startValue < endValue)        {           int midValue = (startValue + endValue) / 2;            int smallerCount = 0;            for (int i = 0; i < n; i++) {            for(int j = 0 ; j < n ; j++)            {            if(matrix[i][j] <= midValue)            {            smallerCount++;            }            }}if (smallerCount < k) {startValue = midValue + 1;} else {endValue = midValue;    }}return startValue;}/** * Kth smallest element in a BST(Binary Search Tree) * @param root * @param k * @return */public int findKthSmallestElementInBST(TreeNode root , int k){if (root == null){return -1;  }int leftSize = getTreeSize(root.left);  if (k == leftSize+1){      return root.value;  }else if (k <= leftSize){      return findKthElementInBST(root.left,k);  }else{      return findKthElementInBST(root.right, k-leftSize-1);  }  }/** * 查找两个有序数组中第k小的数 * @param nums1有序数组1 * @param startIndex1数组1开始位置 * @param nums2有序数组2 * @param startIndex2数组2开始位置 * @param k * @return */public int findKthSmallestElementInTwoSortArray(int nums1[] ,int startIndex1, int nums2[] , int startIndex2 , int k){int m = nums1.length - startIndex1;int n = nums2.length - startIndex2;//确保数组2的长度较大，方便之后的操作if(m > n){return findKthSmallestElementInTwoSortArray(nums2, startIndex2, nums1, startIndex1, k);}if(m == 0){return nums2[k-1];}if(k <= 1){return Math.min(nums1[startIndex1], nums2[startIndex2]);}int minMid = Math.min(k/2,m);int maxMid = k - minMid;if(nums1[minMid + startIndex1 - 1] < nums2[maxMid +startIndex2 -1]){return findKthSmallestElementInTwoSortArray(nums1, startIndex1+minMid, nums2, startIndex2, k - minMid);}else if(nums1[minMid + startIndex1 - 1] > nums2[maxMid+startIndex2 -1]){return findKthSmallestElementInTwoSortArray(nums1, startIndex1, nums2, startIndex2 + maxMid, k - maxMid);}else{return nums1[minMid+startIndex1 -1];}}/** *  * 获取树的大小(节点个数) * @param root根节点 * @return */public int getTreeSize(TreeNode root){  if (root == null){    return 0;}return 1+getTreeSize(root.left) + getTreeSize(root.right);          }  public static void main(String[] args) {int nums1[] = {1,3,5,7,9};int nums2[] = {2,4,6,8,10};int s = new KthSmallestElement().findKthSmallestElementInTwoSortArray(nums1, 0, nums2, 0, 6);System.out.println(s);}}/** * 树节点类 * @author 健身小码哥 * */class TreeNode{TreeNode left;TreeNode right;int value;}