OpenJudge 由中根顺序和后根序列重建二叉树

Posted 2021-01-20 jeffrey-y

【题解】

        后根序列的最后一个元素即为二叉树的树根root。root将中根序列分为两部分，左半边是左子树的中根序列，而右半边则是右部分的中根序列。同时后根序列依照左子树和右子树节点数也可以被分为左子树的后根序列和右子树的后根序列。于是便可依此递归地按左右子树的后根、中根序列重建子树，最终重建二叉树。

【代码】

 1 #include <iostream>
 2 using namespace std;
 3 /*build BT from its inorder and postorder*/
 4 
 5 
 6 constexpr int MAXN = 65540;
 7 int postorder[MAXN], inorder[MAXN];
 8 
 9 struct btn
10 {
11     int data;
12     btn *left;
13     btn *right;
14 };
15 
16 btn *build_tree(int io1, int io2, int po1, int po2)
17 {
18     /*io1, io2 are the begining and ending points of inorder sequence respectively*/
19     /*po1, po2 are the begining and ending points of postorder sequence respectively*/
20     int i = 0, ion = io2 - io1 + 1;
21     btn* root = new btn;
22     root->data = postorder[po2];
23     root->left = NULL;
24     root->right = NULL;
25     for (i = 0; i < ion; i++) {
26         if (root->data == inorder[io1 + i])break;
27     }
28     if (i > 0) root->left = build_tree(io1, io1 + i - 1, po1, po1 + i - 1);
29     if (io1 + i < io2) root->right = build_tree(io1 + i + 1, io2, po1 + i, po2 - 1);
30     return root;
31 }
32 
33 void preorder(btn *root, int vnum, int &count)
34 {
35     if (root != NULL) {
36         cout << root->data;
37         if (count < vnum - 1) {
38             cout << " ";
39             count++;
40             preorder(root->left, vnum, count);
41             preorder(root->right, vnum, count);
42         }
43         else cout << endl;
44     }
45 }
46 
47 void deletetree(btn *root)
48 {
49     if (root != NULL) {
50         delete(root->left);
51         delete(root->right);
52         delete root;
53         root = NULL;
54     }
55 }
56 
57 int main()
58 {
59     int c = 0, i = 0, vnum = 0;
60     while (cin >> inorder[i++]) {
61         if (cin.get() != ‘ ‘)break;
62         /*Alththough cin will skip the blank space, the cursor stays at the blank after cin reads a number，thus cin.get() can fetch the 
63         blank space.*/
64     }
65 
66     vnum = i;
67     i = 0;
68     while (cin >> postorder[i++]) {
69         if (cin.get() != ‘ ‘)break;
70     }
71     btn *root = build_tree(0, i - 1, 0, i - 1);
72     preorder(root, vnum, c);
73     deletetree(root);
74     return 0;
75 }