UVa 839 天平

Posted 2020-08-21 谦谦君子，陌上其华

原题链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=780

先建立二叉树，之后遍历。

 1 #include<iostream>
 2 using namespace std;
 3 
 4 struct Node
 5 {
 6     int W1, D1, W2, D2;
 7     Node *left;
 8     Node *right;
 9 };
10 
11 bool flag;
12 
13 Node* creatTree()
14 {
15     Node* root = new Node;
16     cin >> root->W1 >> root->D1 >> root->W2 >> root->D2;
17     if (root->W1 == 0)  root->left = creatTree();
18     else                root->left = NULL;
19     if (root->W2 == 0)    root->right = creatTree();
20     else                root->right = NULL;
21     return root;
22 }
23 
24 int balance(Node* root)
25 {
26     if (flag == false)  return 0;
27     if (root->W1 == 0)   root->W1 = balance(root->left);
28     if (root->W2 == 0)     root->W2 = balance(root->right);
29     if (root->W1*root->D1 != root->W2*root->D2)  flag = false;
30     return(root->W1 + root->W2);
31 }
32 
33 void remove_tree(Node* u)
34 {
35     if (u == NULL)  return;
36     remove_tree(u->left);
37     remove_tree(u->right);
38     delete u;
39 }
40 
41 int main()
42 {
43     int t;
44     cin >> t;
45     while (t--)
46     {
47         Node* root = NULL;
48         flag = true;
49         root=creatTree();
50         flag = true;
51         balance(root);
52         if (flag == true)  cout << "YES" << endl;
53         else               cout << "NO" << endl;
54         remove_tree(root);
55         if (t != 0)   cout << endl;
56     }
57 }

刘汝佳的书上给出了一个引用传值的简单的方法

 1 #include<iostream>
 2 using namespace std;
 3 
 4 bool solve(int &W)
 5 {
 6     int W1, D1, W2, D2;
 7     bool b1 = true, b2 = true;
 8     cin >> W1 >> D1 >> W2 >> D2;
 9     if (!W1)  b1 = solve(W1);
10     if (!W2)  b2 = solve(W2);
11     W = W1 + W2;
12     return b1&&b2 && (W1*D1 == W2*D2);
13 }
14 
15 int main()
16 {
17     int T, W;
18     cin >> T;
19     while (T--)
20     {
21         if (solve(W))  cout << "YES" << endl;
22         else cout << "NO" << endl;
23         if (T)  cout << endl;
24     }
25     return 0;
26 }