JZ-C-06

 1 //============================================================================
 2 // Name        : JZ-C-06.cpp
 3 // Author      : Laughing_Lz
 4 // Version     :
 5 // Copyright   : All Right Reserved
 6 // Description : Hello World in C++, Ansi-style
 7 //============================================================================
 8 
 9 #include <iostream>
10 #include <stdio.h>
11 #include <iomanip>
12 #include <exception>
13 using namespace std;
14 
15 typedef struct BinaryTreeNode { //二叉树结点
16     int data; //数据域
17     struct BinaryTreeNode *left; //左孩子
18     struct BinaryTreeNode *right; //右孩子
19 } BTN;
20 BTN* ConstructBTCore(int *startPreOrder, int *endPreOrder, int *startInOrder,
21         int *endInOrder); //函数调用在前，定义在后，需要先进行函数原型说明
22 
23 BTN* ConstructBT(int *preOrder, int *inOrder, int length) {
24     if (preOrder == NULL || inOrder == NULL || length <= 0) {
25         cout << "序列为空？请检查两个序列" << endl;
26         return NULL;
27     }
28     BTN *root = ConstructBTCore(preOrder, preOrder + length - 1, inOrder,
29             inOrder + length - 1);
30     cout << root->data << endl;//返回根
31     return root;
32 }
33 
34 BTN* ConstructBTCore(int *startPreOrder, int *endPreOrder, int *startInOrder,
35         int *endInOrder) {
36     BTN *root = new BTN(); //生成结点
37     root->data = startPreOrder[0];
38     root->left = root->right = NULL;
39     if (endPreOrder == startPreOrder) {
40         if (startInOrder == endInOrder && *startPreOrder == *startInOrder) {
41             return root;
42         } else {
43             cout << "前序序列和中序序列不匹配，错误！" << endl;
44             throw std::exception();
45         }
46     }
47     int inOrderLength = endInOrder - startInOrder; //中序遍历序列长度
48     int i = 0;
49     for (; i < inOrderLength + 1; i++) { //在中序遍历序列里查找头结点的值
50 //        int rootInOrder =*startInOrder;
51 //        while(*startInOrder+rootInOrder){}
52         if (*startPreOrder == *(startInOrder + i)) { //和前序遍历序列的头结点比较
53             root->data = *startPreOrder; //头结点
54             if (i > 0) { //如果有左子树
55                 root->left = ConstructBTCore(startPreOrder + 1,
56                         startPreOrder + i, startInOrder, startInOrder + i - 1); //递归遍历左子树序列
57             }
58             if (i < inOrderLength) { //如果有右子树
59                 root->right = ConstructBTCore(startPreOrder + i + 1,
60                         endPreOrder, startInOrder + i + 1, endInOrder); //递归遍历右子树序列
61             }
62             break; //退出循环
63         }
64     }
65     if (i == inOrderLength + 1) { //说明中序遍历序列中没有和前序遍历序列头结点值相同的结点，错误
66         cout << "中序序列中没有前序序列的头结点，错误！" << endl;
67         throw std::exception();
68     }
69     return root;
70 }
71 int main() {
72     const int length = 7; //完全二叉树
73     int preorder[length] = { 1, 2, 4, 5, 3, 6, 7 };
74     int inorder[length] = { 4, 2, 5, 1, 6, 3, 7 };
75 //    const int length = 8;//普通二叉树
76 //    int preorder[length] = { 1, 2, 4, 7, 3, 5, 6, 8 };
77 //    int inorder[length] = { 4, 7, 2, 1, 5, 3, 8, 6 };
78 //  const int length = 7;//不匹配的序列
79 //  int preorder[length] = {1, 2, 4, 5, 3, 6, 7};
80 //  int inorder[length] = {4, 2, 8, 1, 6, 3, 7};
81 //    const int length = 5;//所有结点都没有右子结点
82 //    int preorder[length] = { 1, 2, 3, 4, 5 };
83 //    int inorder[length] = { 5, 4, 3, 2, 1 };
84 //    const int length = 5;//    所有结点都没有左子结点
85 //    int preorder[length] = {1, 2, 3, 4, 5};
86 //    int inorder[length] = {1, 2, 3, 4, 5};
87 //    const int length = 1;//树中只有一个结点
88 //    int preorder[length] = {1};
89 //    int inorder[length] = {1};
90     ConstructBT(preorder, inorder, length);
91 //    ConstructBT(NULL, NULL, 0);//输入空指针
92 
93     return 0;
94 }