题目：
给定一棵二叉搜索树，请找出其中第k大的结点。
分析：
对二叉搜索树的中序遍历结果，是一个从小到大的有序序列，要找第K大结点，可以将有序序列变成从大到小的，也就是使用右根左的顺序递归遍历，同时使用一个计数器，当计数器达到k的时候，就是第k大结点了。
解法：
package com.wsy;
class Tree {
private int value;
private Tree left;
private Tree right;
public Tree() {
}
public Tree(int value) {
this.value = value;
this.left = this.right = null;
}
public Tree(int value, Tree left, Tree right) {
this.value = value;
this.left = left;
this.right = right;
}
public int getValue() {
return value;
}
public void setValue(int value) {
this.value = value;
}
public Tree getLeft() {
return left;
}
public void setLeft(Tree left) {
this.left = left;
}
public Tree getRight() {
return right;
}
public void setRight(Tree right) {
this.right = right;
}
}
public class Main {
public static int count;
public static void main(String[] args) {
Tree tree = init();
int k = 3;
getKth(tree, k);
}
public static Tree init() {
Tree tree8 = new Tree(8);
Tree tree6 = new Tree(6);
Tree tree4 = new Tree(4);
Tree tree2 = new Tree(2);
Tree tree7 = new Tree(7, tree6, tree8);
Tree tree3 = new Tree(3, tree2, tree4);
Tree tree5 = new Tree(5, tree3, tree7);
return tree5;
}
public static void getKth(Tree tree, int k) {
if (tree == null || k < 1) {
return;
}
if (tree.getRight() != null) {
getKth(tree.getRight(), k);
}
count++;
if (count == k) {
System.out.println("第" + k + "大的结点是：" + tree.getValue());
}
if (tree.getLeft() != null) {
getKth(tree.getLeft(), k);
}
}
}