T₀ = 0, T₁ = 1, T₂ = 1, and T_n+3 = T_n + T_n+1 + T_n+2 for n >= 0.

Given n, return the value of T_n.

Example 1:

Input: n = 4
Output: 4
Explanation:
T_3 = 0 + 1 + 1 = 2
T_4 = 1 + 1 + 2 = 4

Example 2:

Input: n = 25
Output: 1389537

Constraints:

• 0 <= n <= 37
• The answer is guaranteed to fit within a 32-bit integer, ie. answer <= 2^31 - 1.

Solution: DP

Time complexity: O(n)
Space complexity: O(n) -> O(1)

// Author: Huahua
class Solution {
public:
  int tribonacci(int n) {
    if (n == 0) return n;
    int t0 = 0;
    int t1 = 1;
    int t2 = 1;
    int t = 1;
    for (int i = 3; i <= n; ++i) {
      t = t0 + t1 + t2;
      t0 = t1;
      t1 = t2;
      t2 = t;
    }
    return t;
  }
};

The post 花花酱 LeetCode 1137. N-th Tribonacci Number appeared first on Huahua's Tech Road.

Given a binary tree, return the inorder traversal of its nodes’ values.

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

Solution: Recursion

Time complexity: O(n)

Space complexity: O(h)

C++

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  vector<int> inorderTraversal(TreeNode* root) {    
    vector<int> ans;
    inorderTraversal(root, ans);
    return ans;
  }
private:
  void inorderTraversal(TreeNode* root, vector<int>& ans) {
    if (root == nullptr) return;
    inorderTraversal(root->left, ans);
    ans.push_back(root->val);
    inorderTraversal(root->right, ans);    
  }
};

Solution 2: Iterative

// Author: Huahua
// Running time: 3 ms
lass Solution {
public:
  vector<int> inorderTraversal(TreeNode* root) {
    if (root == nullptr) return {};
    vector<int> ans;
    stack<TreeNode*> s;
    TreeNode* curr = root;
    while (curr != nullptr || !s.empty()) {
      while (curr != nullptr) {
        s.push(curr);
        curr = curr->left;
      }
      curr = s.top(); s.pop();
      ans.push_back(curr->val);
      curr = curr->right;
    }    
    return ans;
  }
};

The post 花花酱 LeetCode 94. Binary Tree Inorder Traversal appeared first on Huahua's Tech Road.

Problem:

Given a binary tree, return the postorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},

1
    \
     2
    /
   3

return [3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

Solution 1:

// Author: Huahua
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> ans;        
        postorderTraversal(root, ans);
        return ans;
    }
    
    void postorderTraversal(TreeNode* root, vector<int>& ans) {
        if (!root) return;
        postorderTraversal(root->left, ans);
        postorderTraversal(root->right, ans);
        ans.push_back(root->val);
    }
};

Solution 2:

// Author: Huahua
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        if (!root) return {};
        vector<int> ans;
        const vector<int> l = postorderTraversal(root->left);
        const vector<int> r = postorderTraversal(root->right);
        ans.insert(ans.end(), l.begin(), l.end());
        ans.insert(ans.end(), r.begin(), r.end());
        ans.push_back(root->val);
        return ans;
    }
};

Solution 3:

// Author: Huahua
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        if (!root) return {};
        deque<int> ans;
        stack<TreeNode*> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode* n = s.top();
            s.pop();
            ans.push_front(n->val); // O(1)
            if (n->left) s.push(n->left);
            if (n->right) s.push(n->right);
        }   
        return vector<int>(ans.begin(), ans.end());
    }
};

The post 花花酱 LeetCode 145. Binary Tree Postorder Traversal appeared first on Huahua's Tech Road.