main.cpp

// Source: https://leetcode.com/problems/intersection-of-two-linked-lists
// Title: Intersection of Two Linked Lists
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// Given the heads of two singly linked-lists `headA` and `headB`, return the node at which the two lists intersect. If the two linked lists have no intersection at all, return `null`.
//
// For example, the following two linked lists begin to intersect at node `c1`:
// https://assets.leetcode.com/uploads/2021/03/05/160_statement.png
// The test cases are generated such that there are no cycles anywhere in the entire linked structure.
//
// **Note** that the linked lists must **retain their original structure** after the function returns.
//
// **Custom Judge:**
//
// The inputs to the **judge** are given as follows (your program is **not** given these inputs):
//
// - `intersectVal` - The value of the node where the intersection occurs. This is `0` if there is no intersected node.
// - `listA` - The first linked list.
// - `listB` - The second linked list.
// - `skipA` - The number of nodes to skip ahead in `listA` (starting from the head) to get to the intersected node.
// - `skipB` - The number of nodes to skip ahead in `listB` (starting from the head) to get to the intersected node.
//
// The judge will then create the linked structure based on these inputs and pass the two heads, `headA` and `headB` to your program. If you correctly return the intersected node, then your solution will be **accepted**.
//
// **Example 1:**
// https://assets.leetcode.com/uploads/2021/03/05/160_example_1_1.png
//
// ```
// Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
// Output: Intersected at '8'
// Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect).
// From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.
// - Note that the intersected node's value is not 1 because the nodes with value 1 in A and B (2^nd node in A and 3^rd node in B) are different node references. In other words, they point to two different locations in memory, while the nodes with value 8 in A and B (3^rd node in A and 4^th node in B) point to the same location in memory.
// ```
//
// **Example 2:**
// https://assets.leetcode.com/uploads/2021/03/05/160_example_2.png
//
// ```
// Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
// Output: Intersected at '2'
// Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect).
// From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.
// ```
//
// **Example 3:**
// https://assets.leetcode.com/uploads/2021/03/05/160_example_3.png
//
// ```
// Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
// Output: No intersection
// Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
// Explanation: The two lists do not intersect, so return null.
// ```
//
// **Constraints:**
//
// - The number of nodes of `listA` is in the `m`.
// - The number of nodes of `listB` is in the `n`.
// - `1 <= m, n <= 3 * 10^4`
// - `1 <= Node.val <= 10^5`
// - `0 <= skipA <= m`
// - `0 <= skipB <= n`
// - `intersectVal` is `0` if `listA` and `listB` do not intersect.
// - `intersectVal == listA[skipA] == listB[skipB]` if `listA` and `listB` intersect.
//
// **Follow up:** Could you write a solution that runs in `O(m + n)` time and use only `O(1)` memory?
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

#include <stack>
#include <vector>

using namespace std;

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) {}
};

// Use DFS (Recursion)
class Solution {
 public:
  vector<int> preorderTraversal(TreeNode *root) {
    auto ans = vector<int>();
    _preorder(root, ans);
    return ans;
  }

 private:
  void _preorder(TreeNode *node, vector<int> &ans) {
    if (!node) return;
    ans.push_back(node->val);
    _preorder(node->left, ans);
    _preorder(node->right, ans);
  }
};

// Use DFS (Stack)
class Solution2 {
 public:
  vector<int> preorderTraversal(TreeNode *root) {
    if (!root) return {};

    auto ans = vector<int>();
    auto st = stack<TreeNode *>();
    st.push(root);

    while (!st.empty()) {
      auto node = st.top();
      st.pop();
      ans.push_back(node->val);
      if (node->right) st.push(node->right);
      if (node->left) st.push(node->left);
    }

    return ans;
  }
};