本题要求给定二叉树的高度。
函数接口定义:
int GetHeight( BinTree BT );
其中BinTree
结构定义如下:
typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
ElementType Data;
BinTree Left;
BinTree Right;
};
要求函数返回给定二叉树BT的高度值。
裁判测试程序样例:
#include <stdio.h>
#include <stdlib.h>
typedef char ElementType;
typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
ElementType Data;
BinTree Left;
BinTree Right;
};
BinTree CreatBinTree(); /* 实现细节忽略 */
int GetHeight( BinTree BT );
int main()
{
BinTree BT = CreatBinTree();
printf("%d\n", GetHeight(BT));
return 0;
}
/* 你的代码将被嵌在这里 */
输出样例(对于图中给出的树):
4
思路这个题目就是遍历左子树和右子树,也就是遍历每一层,看看总的层数有多少。
第一次的代码,样例五不能过
int GetHeight( BinTree BT ){
//left
int left_height = 0, right_height = 0;
if(BT->Left != NULL)
left_height = 1 + GetHeight(BT->Left);
else
left_height = 1;
//right
if(BT->Right != NULL)
right_height = 1 + GetHeight(BT->Right);
else
right_height = 1;
if(left_height > right_height)
return left_height;
else return right_height;
}
稍微优化一下
int GetHeight( BinTree BT ){
if(BT == NULL)
return 0;
//left
int left_height = 0, right_height = 0;
left_height = 1 + GetHeight(BT->Left);
//right
right_height = 1 + GetHeight(BT->Right);
if(left_height > right_height)
return left_height;
else return right_height;
}