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;}
};