# 二叉搜索树的第 k 个结点

## 题目：给定一棵二叉搜索树，请找出其中的第k大的结点。

### 结点定义

private static class BinaryTreeNode {
private int val;
private BinaryTreeNode left;
private BinaryTreeNode right;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
@Override
public String toString() {
return val + "";
}
}

### 代码实现

public class Test63 {
private static class BinaryTreeNode {
private int val;
private BinaryTreeNode left;
private BinaryTreeNode right;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
@Override
public String toString() {
return val + "";
}
}
public static BinaryTreeNode kthNode(BinaryTreeNode root, int k) {
if (root == null || k < 1) {
return null;
}
int[] tmp = {k};
return kthNodeCore(root, tmp);
}
private static BinaryTreeNode kthNodeCore(BinaryTreeNode root, int[] k) {
BinaryTreeNode result = null;
// 先成左子树中找
if (root.left != null) {
result =  kthNodeCore(root.left, k);
}
// 如果在左子树中没有找到
if (result == null) {
// 说明当前的根结点是所要找的结点
if(k[0] == 1) {
result = root;
} else {
// 当前的根结点不是要找的结点，但是已经找过了，所以计数器减一
k[0]--;
}
}
// 根结点以及根结点的右子结点都没有找到，则找其右子树
if (result == null && root.right != null) {
result = kthNodeCore(root.right, k);
}
return result;
}
public static void main(String[] args) {
BinaryTreeNode n1 = new BinaryTreeNode(1);
BinaryTreeNode n2 = new BinaryTreeNode(2);
BinaryTreeNode n3 = new BinaryTreeNode(3);
BinaryTreeNode n4 = new BinaryTreeNode(4);
BinaryTreeNode n5 = new BinaryTreeNode(5);
BinaryTreeNode n6 = new BinaryTreeNode(6);
BinaryTreeNode n7 = new BinaryTreeNode(7);
BinaryTreeNode n8 = new BinaryTreeNode(8);
BinaryTreeNode n9 = new BinaryTreeNode(9);
n1.left = n2;
n1.right = n3;
n2.left = n4;
n2.right = n5;
n3.left = n6;
n3.right = n7;
n4.left = n8;
n4.right = n9;
print(n1);
System.out.println();
for (int i = 0; i <= 10; i++) {
System.out.printf(kthNode(n1, i) + ", ");
}
}
/**
* 中序遍历一棵树
* @param root
*/
private static void print(BinaryTreeNode root) {
if (root != null) {
print(root.left);
System.out.printf("%-3d", root.val);
print(root.right);
}
}
}