C++_学习随笔_重构二叉树和判断是否是对称二叉树

 1 #include<iostream>
 2 #include<vector>
 3 using namespace std;
 4 
 5 struct TreeNode {
 6     int val;
 7     TreeNode* left;
 8     TreeNode* right;
 9     TreeNode(int x):val(x),left(NULL),right(NULL){}
10 };
11 
12 //根据前序遍历和中序遍历重构二叉树
13 TreeNode* reConstructBinaryTree(vector<int>pre, vector<int>vin) {
14     if (pre.empty() || vin.empty())
15         return nullptr;
16     //根节点
17     TreeNode *root = new TreeNode(pre[0]);
18     int root_index = 0;
19     //找到根节点的索引值
20     for (int i = 0; i < pre.size(); i++) {
21         if (vin[i] == pre[0]) {
22             root_index = i;
23             break;
24         }
25     }
26     vector<int> pre_left, pre_right, vin_left, vin_right;
27     for (int i = 0; i < root_index; i++) {
28         pre_left.push_back(pre[i + 1]);
29         vin_left.push_back(vin[i]);
30     }
31 
32     for (int i = root_index + 1; i < pre.size(); i++) {
33         pre_right.push_back(pre[i]);
34         vin_right.push_back(vin[i]);
35     }
36 
37     root->left = reConstructBinaryTree(pre_left, vin_left);
38     root->right = reConstructBinaryTree(pre_right, vin_right);
39 
40     return root;
41 }
42 
43 
44 //判断是否是对称的
45 bool isSymmetrical(TreeNode* p1, TreeNode* p2) {
46     if (p1 == NULL && p2 == NULL)
47         return true;
48     if (p1 == NULL || p2 == NULL)
49         return false;
50     if (p1->val != p2->val)
51         return false;
52     return isSymmetrical(p1->left, p2->right) && isSymmetrical(p1->right, p2->left);
53 }
54 
55 
56 bool isSymmetrical(TreeNode* pRoot) {
57     if (pRoot == NULL) {
58         return true;
59     }
60     return isSymmetrical(pRoot, pRoot);
61 }
62 
63 
64 int main() {
65     vector<int> pre, vin;
66     cout << "请输入节点数："<<endl;
67     int n = 0,temp;
68     cin >> n;
69     cout << "请输入前序遍历：" << endl;
70     for (int i = 0; i < n; i++) {
71         cin >> temp;
72         pre.push_back(temp);
73     }
74     cout << "请输入中序遍历：" << endl;
75     for (int i = 0; i < n; i++) {
76         cin >> temp;
77         vin.push_back(temp);
78     }
79     bool is;
80     TreeNode* root_node = reConstructBinaryTree(pre, vin);
81     is = isSymmetrical(root_node);
82     cout << is;
83     
84 }