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二叉树的前、中、后序遍历【c++】

news 2025/7/28 9:48:43

前序遍历：根左右
中序遍历：左根右
后序遍历：左右根

#include <iostream>
#include <vector>
using namespace std;//双链表节点结构
typedef struct treeNode {int value;struct treeNode* left;struct treeNode* right;treeNode(int x) : value(x), left(nullptr), right(nullptr) {}
} Node;//前序遍历
void preOrder(Node* root, vector<int> &allval)
{if (!root){return;}allval.push_back(root->value);  //遍历根节点preOrder(root->left, allval);   //遍历左节点preOrder(root->right, allval);  //遍历右节点
}//中序遍历
void inOrder(Node* root, vector<int> &allval)
{if (!root){return;}inOrder(root->left, allval);   //遍历左节点allval.push_back(root->value);  //遍历根节点inOrder(root->right, allval);  //遍历右节点
}//后序遍历
void postOrder(Node* root, vector<int> &allval)
{if (!root){return;}postOrder(root->left, allval);   //遍历左节点postOrder(root->right, allval);  //遍历右节点allval.push_back(root->value);  //遍历根节点
} int main()
{Node* root = new Node(1);root->left = new Node(2);root->right = new Node(3);root->left->left = new Node(4);root->left->right = new Node(5);//         1//      2     3//   4    5cout << "前序遍历结果：" << endl;  //12453vector<int> ves;preOrder(root, ves);for (int i = 0; i < ves.size(); i++){cout << ves.at(i) << " ";}cout << endl;cout << "中序遍历结果：" << endl;  //42513vector<int> ves1;inOrder(root, ves1);for (int i = 0; i < ves1.size(); i++){cout << ves1.at(i) << " ";}cout << endl;cout << "后序遍历结果：" << endl;  //45231vector<int> ves2;postOrder(root, ves2);for (int i = 0; i < ves2.size(); i++){cout << ves2.at(i) << " ";}cout << endl;system("pause");return 0;
}