6-11 先序輸出葉結點 (15 分) 數據結構與算法題目集（中文）

本題要求按照先序遍歷的順序輸出給定二叉樹的葉結點。

函數接口定義：

void PreorderPrintLeaves( BinTree BT );

其中BinTree結構定義如下：

typedef struct TNode *Position;
typedef Position BinTree;
struct TNode{
    ElementType Data;
    BinTree Left;
    BinTree Right;
};

函數PreorderPrintLeaves應按照先序遍歷的順序輸出給定二叉樹BT的葉結點，格式為一個空格跟着一個字符。

裁判測試程序樣例：

#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(); /* 實現細節忽略 */
void PreorderPrintLeaves( BinTree BT );

int main()
{
    BinTree BT = CreatBinTree();
    printf("Leaf nodes are:");
    PreorderPrintLeaves(BT);
    printf("\n");

    return 0;
}
/* 你的代碼將被嵌在這里 */

輸出樣例（對於圖中給出的樹）：

Leaf nodes are: D E H I

復習考研，書上只寫了中序遍歷的非遞歸算法，所以找一道題試試前序遍歷能不能寫出來。

代碼：

#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 Insert(BinTree T,ElementType X) {
    if(T == NULL) {
        T = (BinTree)malloc(sizeof(TNode));
        T -> Left = T -> Right = NULL;
        T -> Data = X;
    }
    else if(X < T -> Data) {
        T -> Left = Insert(T -> Left,X);
    }
    else {
        T -> Right = Insert(T -> Right,X);
    }
    return T;
}
BinTree CreatBinTree();
void PreorderPrintLeaves( BinTree BT );

int main()
{
    BinTree BT = CreatBinTree();
    printf("Leaf nodes are:");
    PreorderPrintLeaves(BT);
    printf("\n");

    return 0;
}
BinTree CreatBinTree() {
    BinTree Bt = NULL;
    int n;
    ElementType ch;
    scanf("%d",&n);
    for(int i = 0;i < n;i ++) {
        getchar();
        scanf("%c",&ch);
        Bt = Insert(Bt,ch);
    }
    return Bt;
}
void PreorderPrintLeaves( BinTree BT ) {
    BinTree b[100];
    int c = 0;
    if(BT) b[c ++] = BT;
    while(c != 0) {
        BinTree temp = b[-- c];
        if(!(temp -> Left || temp -> Right)) printf(" %c",temp -> Data);
        if(temp -> Right) b[c ++] = temp -> Right;
        if(temp -> Left) b[c ++] = temp -> Left;
    }
}