23. 合并 K 个升序链表 - 力扣（LeetCode）

若lists有k个元素，调用k - 1次（两个有序链表的合并）即可

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* _mergeKLists(ListNode *l1, ListNode *l2)
    {
        ListNode *head = new ListNode;
        ListNode *del = head;
        while (l1 && l2)
        {
            if (l1->val < l2->val) head->next = l1, l1 = l1->next;
            else head->next = l2, l2 = l2->next;
            head = head->next;
        }
        if (l1) head->next = l1;
        else head->next = l2;
        ListNode *res = del->next;
        delete del;
        return res;
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.size() == 0) return nullptr;
        ListNode *ans = lists[0];
        for (int i = 1; i < lists.size(); ++ i)
        {
            ans = _mergeKLists(ans, lists[i]);
        }
        return ans;
    }
};

94. 二叉树的中序遍历 - 力扣（LeetCode）

使用栈模拟递归调用的过程

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:

    vector<int> inorderTraversal(TreeNode* root) {
        stack<TreeNode *> st;
        vector<int> ans;
        while (root || st.size())
        {
            while (root)
            {
                st.push(root);
                root = root->left;
            }
            root = st.top(); st.pop();
            ans.push_back(root->val);
            root = root->right;
        }
        return ans;
    }
};