非递归版本：
 

#include<iostream>
#include<stack>
using namespace std;
struct TreeNode {
    int _val;
    TreeNode* left, * right;
    TreeNode(const int& val) :_val(val), left(nullptr), right(nullptr) {}
    ~TreeNode() {
        delete left, right;
    }
};
void inorderTraversal(TreeNode* root) {
    TreeNode* cur = root;
    stack<TreeNode*> st;
    while (cur || !st.empty()) {
        while (cur) {
            st.push(cur);
            cur = cur->left;
        }
        cur = st.top();
        st.pop();
        cout << cur->_val << " ";
        cur = cur->right;
    }
}
int main() {
    // 用例1：简单的二叉搜索树
    TreeNode* root1 = new TreeNode(5);
    root1->left = new TreeNode(3);
    root1->right = new TreeNode(7);
    root1->left->left = new TreeNode(2);
    root1->left->right = new TreeNode(4);
    root1->right->left = new TreeNode(6);
    root1->right->right = new TreeNode(8);
    cout << "Inorder traversal of BST: ";
    inorderTraversal(root1);
    cout << endl;

    // 用例2：只有一个节点的树
    TreeNode* root2 = new TreeNode(10);
    cout << "Inorder traversal of single node tree: ";
    inorderTraversal(root2);
    cout << endl;

    // 用例3：空树
    TreeNode* root3 = nullptr;
    cout << "Inorder traversal of empty tree: ";
    inorderTraversal(root3);
    cout << endl;

    // 用例4：不平衡的树
    TreeNode* root4 = new TreeNode(1);
    root4->left = new TreeNode(2);
    root4->left->left = new TreeNode(3);
    root4->left->left->left = new TreeNode(4);
    cout << "Inorder traversal of unbalanced tree: ";
    inorderTraversal(root4);
    cout << endl;

    // 用例5：退化成链表的树
    TreeNode* root5 = new TreeNode(1);
    TreeNode* cur = root5;
    for (int i = 2; i <= 5; ++i) {
        cur->right = new TreeNode(i);
        cur = cur->right;
    }
    cout << "Inorder traversal of linked list tree: ";
    inorderTraversal(root5);
    cout << endl;

    // 释放所有节点的内存
    delete root1;
    delete root2;
    delete root3; // 虽然是空指针，但delete空指针是安全的
    delete root4;
    delete root5;

    return 0;
}

值得注意的点：

1.stack中存放的是树的节点的地址也就是TreeNode*类型。

2. TreeNode的析构函数千万不要忘记，因为我们是在堆上申请的内存。

3.delete空指针什么都不做。

下面我们实现递归版本，这个要容易地多。

#include<iostream>
#include<stack>
using namespace std;
struct TreeNode {
    int _val;
    TreeNode* left, * right;
    TreeNode(const int& val) :_val(val), left(nullptr), right(nullptr) {}
    ~TreeNode() {
        delete left, right;
    }
};
void inorderTraversal(TreeNode* root) {
    if (root == nullptr)return;
    inorderTraversal(root->left);
    cout << root->_val << " ";
    inorderTraversal(root->right);    
}
int main() {
    // 用例1：简单的二叉搜索树
    TreeNode* root1 = new TreeNode(5);
    root1->left = new TreeNode(3);
    root1->right = new TreeNode(7);
    root1->left->left = new TreeNode(2);
    root1->left->right = new TreeNode(4);
    root1->right->left = new TreeNode(6);
    root1->right->right = new TreeNode(8);
    cout << "Inorder traversal of BST: ";
    inorderTraversal(root1);
    cout << endl;

    // 用例2：只有一个节点的树
    TreeNode* root2 = new TreeNode(10);
    cout << "Inorder traversal of single node tree: ";
    inorderTraversal(root2);
    cout << endl;

    // 用例3：空树
    TreeNode* root3 = nullptr;
    cout << "Inorder traversal of empty tree: ";
    inorderTraversal(root3);
    cout << endl;

    // 用例4：不平衡的树
    TreeNode* root4 = new TreeNode(1);
    root4->left = new TreeNode(2);
    root4->left->left = new TreeNode(3);
    root4->left->left->left = new TreeNode(4);
    cout << "Inorder traversal of unbalanced tree: ";
    inorderTraversal(root4);
    cout << endl;

    // 用例5：退化成链表的树
    TreeNode* root5 = new TreeNode(1);
    TreeNode* cur = root5;
    for (int i = 2; i <= 5; ++i) {
        cur->right = new TreeNode(i);
        cur = cur->right;
    }
    cout << "Inorder traversal of linked list tree: ";
    inorderTraversal(root5);
    cout << endl;

    // 释放所有节点的内存
    delete root1;
    delete root2;
    delete root3; // 虽然是空指针，但delete空指针是安全的
    delete root4;
    delete root5;

    return 0;
}