（二叉搜索树6-12）The Kth Largest X in BST--C语言

Given a binary search tree T, you are supposed to find the node with key value X, and return K if X is the K-th largest key.

Format of function:

int KthLargest ( BinTree T, int X );

where BinTree is defined as the following:

typedef struct TNode *BinTree;
struct TNode{
int Key;
BinTree Left;
BinTree Right;
};

Here T is not empty and all its keys are distinct positive integers. If X is not found, return 0 instead.

Sample program of judge:

#include <stdio.h>
#include <stdlib.h>

typedef struct TNode *BinTree;
struct TNode{
    int Key;
    BinTree Left;
    BinTree Right;
};

BinTree BuildTree(); /* details omitted */
int KthLargest ( BinTree T, int X );

int main()
{
    BinTree T;
    int X;

    T = BuildTree();
    scanf("%d", &X);
    printf("%d", KthLargest(T, X));

    return 0;
}
/* Your function will be put here */

Sample Input 1:

5

Sample Output 1:

4

Sample Input 2:

15

Sample Output 2:

0

Analysis:

用中序遍历得到序列，因为是二叉搜索树，则该序列一定是递增的，所以直接在序列中找输入的数，如果找到了直接返回（count-下标），count是序列的总数，因为序列从0开始；如果没找到直接返回0。

Notice：

• 如果不知道怎么用中序遍历得到序列，可以去看是否二叉搜索树，在这篇文章中，我具体分析了中序遍历得到序列。

Code:

int KthLargest ( BinTree T, int X )
{
    int flag=0;
    BinTree BT=T; 
    BinTree a[100]={NULL},b[100]={NULL};
    int ai=0,bi=0,i;
    while(BT||ai!=0)
    {
        while(BT)
        {
            a[ai++]=BT;
            BT=BT->Left;
        }
        if(a[ai-1]==NULL)
            ai--;
        int top=--ai;
        b[bi++]=a[top];
        BT=a[top];
        a[top]=NULL;
        BT=BT->Right;
    }
    for(i=0;b[i]!=NULL;i++)
    {
        if(b[i]->Key==X)
        {
            flag=1;
            break;
        }
    }
    if(flag==0) return 0;
    else return bi-i;
}