Path Sum III in C++

Suppose we have given a binary tree in which each node holds an integer key. We have to find the paths that sum to a given value. The path should start from root to leaf. We have to find the path where the sum is same.

If the tree is like [5,4,8,11,null,13,4,7,2,null,null,5,1], and sum is 22, then it will be −

The paths are [[5,4,11,2],[5,8,4,5]].

To solve this, we will follow these steps −

• Use the dfs function to solve this problem, the dfs is slightly modified, this will work as follows. This function will take the root, sum, and one temp array
• if root is not present, then return
• if left of root and right of root are empty, then
  • if sum = value of root, then
    • insert value of root into temp, insert temp into result, and delete the last node from the temp
  • return
• insert value of root into the temp
• dfs(left of root, sum – value of root, temp)
• dfs(right of root, sum – value of root, temp)
• delete the last element from temp

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<int> > v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << "[";
      for(int j = 0; j <v[i].size(); j++){
         cout << v[i][j] << ", ";
      }
      cout << "],";
   }
   cout << "]"<<endl;
}
class TreeNode{
   public:
      int val;
      TreeNode *left, *right;
      TreeNode(int data){
         val = data;
         left = right = NULL;
      }
};
void insert(TreeNode **root, int val){
      queue<TreeNode*> q;
      q.push(*root);
      while(q.size()){
         TreeNode *temp = q.front();
         q.pop();
         if(!temp->left){
            if(val != NULL)
               temp->left = new TreeNode(val);
            else
               temp->left = new TreeNode(0);
            return;
      } else {
         q.push(temp->left);
      }
      if(!temp->right){
         if(val != NULL)
         temp->right = new TreeNode(val);
         else
            temp->right = new TreeNode(0);
         return;
      } else {
         q.push(temp->right);
      }
   }
}
TreeNode *make_tree(vector<int> v){
   TreeNode *root = new TreeNode(v[0]);
   for(int i = 1; i<v.size(); i++){
      insert(&root, v[i]);
   }
   return root;
}
class Solution {
   public:
   vector < vector <int> > res;
   void dfs(TreeNode* root, int sum, vector <int>& temp){
      if(!root)return;
      if(!root->left && !root->right){
         if(sum == root->val){
            temp.push_back(root->val);
            res.push_back(temp);
            temp.pop_back();
         }
         return;
      }
      temp.push_back(root->val);
      dfs(root->left, sum - root->val, temp);
      dfs(root->right, sum - root->val, temp);
      temp.pop_back();
   }
   vector<vector<int>> pathSum(TreeNode* root, int sum) {
      res.clear();
      vector <int> temp;
      dfs(root, sum, temp);
      return res;
   }
};
main(){
   Solution ob;
   vector<int> v = {5,4,8,11,NULL,13,4,7,2,NULL,NULL,NULL,NULL,5,1};
   TreeNode *root = make_tree(v);
   print_vector(ob.pathSum(root, 22));
}

Input

[5,4,8,11,null,13,4,7,2,null,null,5,1]
22

Output

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