PAT 2024年春季甲级 A-3 Degree of Skewness（二叉树的构建）

给出后序和中序遍历序列，求出只有左子树和只有右子树的结点之差

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n;
 4 int nl = 0, nr =0;
 5 struct node{
 6     int data, lchild = -1, rchild = -1;
 7 };
 8 vector<node> nodes(1005);
 9 vector<int> in_order, post_order;
10 // 段错误代码，不知为何
11 // string in_order,post_order;
12 // vector<char> tree(1000, ' ');
13 // void rebuild(int p, char root, string in) {
14 //     tree[p] = root;
15 //     if (in.length() <= 1){
16 //         return ;
17 //     }
18 //     string in_left = in.substr(0, in.find(root));
19 //     string in_right = in.substr(in.find(root) + 1);
20 //     for (int i = n - 1; i >= 0; -- i) {
21 //         if (in_left.find(post_order[i])) {
22 //             rebuild(p * 2, post_order[i], in_left);
23 //             break;
24 //         }
25 //     }
26 //     for (int i = n - 1; i >= 0; -- i) {
27 //         if (in_right.find(post_order[i])) {
28 //             rebuild(p * 2 + 1, post_order[i], in_right);
29 //             break;
30 //         }
31 //     }
32 // }
33 int rebuild(int il, int ir, int pl, int pr) { 
34     if (il > ir || pl > pr) {
35         return -1;
36     }
37     int root = post_order[pr];
38     int index;
39     for (index = il; index <= ir; ++ index) {
40         if (in_order[index] == root) {
41             break;
42         }
43     }
44     nodes[root].lchild = rebuild(il, index - 1, pl, pl + index - il - 1);
45     nodes[root].rchild = rebuild(index + 1, ir, pl + index - il, pr - 1);
46     return root;
47 }
48 void dfs(int p) {
49     bool hasleft = false, hasright = false;
50     if (nodes[p].lchild != -1) {
51         hasleft = true;
52         dfs(nodes[p].lchild);
53     }
54     if (nodes[p].rchild != -1) {
55         hasright = true;
56         dfs(nodes[p].rchild);
57     }
58     if (hasleft && !hasright) {
59         nl ++;
60     }
61     if (!hasleft && hasright) {
62         nr ++;
63     }
64 }
65 int main() {
66     cin >> n;
67     int temp;
68     for (int i = 0; i < n; ++ i) {
69         cin >> temp;
70         post_order.push_back(temp);
71     }
72     for (int i = 0; i < n; ++ i) {
73         cin >> temp;
74         in_order.push_back(temp);
75     }
76     int root = rebuild(0, n-1, 0, n-1);
77     dfs(root);
78     cout << root << endl;
79     cout << nl - nr << " = " << nl << " - " << nr;
80 
81 }
82

posted on 2024-12-09 20:00 Coder何阅读(20) 评论(0) 收藏举报

PAT 2024年春季 甲级 A-3 Degree of Skewness（二叉树的构建）

PAT 2024年春季甲级 A-3 Degree of Skewness（二叉树的构建）