LeetCode(Path Sum and Path Sum 2)

제목 요구사항:

Path Sum

 
Total Accepted: 10595 
Total Submissions: 35321 My Submissions
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:
Given the below binary tree and  sum = 22 ,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path  5->4->11->2  which sum is 22.
간단한 귀속 판단:
코드:

  bool hasPathSum(TreeNode *root, int sum) {
    if(root == NULL)
      return false;
    if(root->left == NULL && root->right == NULL)
    {
      if(sum == root->val)
        return true;
      else 
        return false;
    }
    else 
    {
      return hasPathSum(root->left, sum - root->val) || 
          hasPathSum(root->right, sum - root->val);
    }
  }

모든 경로 구하기:
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and  sum = 22
,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

귀속 코드:

vector<vector<int> > pathSum(TreeNode *root, int sum) 
  {
    vector<vector<int> > ans;
    if(root == NULL)
      return ans;
    vector<int> path;
    DFS(root, sum, path, ans);
    return ans;
  }
  
  void DFS(TreeNode* root, int sum, vector<int>& path, vector<vector<int> >& ans)
  {
    if(root->left == NULL && root->right == NULL && sum == root->val)
    {
      path.push_back(root->val);
      ans.push_back(path);
      return;
    }
    path.push_back(root->val);
    if(root->left != NULL)
    {
      DFS(root->left, sum - root->val, path, ans);
      path.pop_back();
    }
    if(root->right != NULL)
    {
      DFS(root->right, sum- root->val, path, ans);
      path.pop_back();
    }
  }