Day-8 Path Sum III

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
    int result=0;
    public int pathSum(TreeNode root, int sum) {
        dfs(root,sum);
        return result;
    }
    
    void dfs(TreeNode node,int sum){
        if(node==null){
            return;
        }
        
        sum(node,sum);
        dfs(node.left,sum);
        dfs(node.right,sum);
    }
    
    void sum(TreeNode n,int s){
 
        if(n==null){
            return;
        }
        int temp=s-n.val;
        if(temp==0){
            result++;
        }
        sum(n.left,temp);
        sum(n.right,temp);
    }
}

class Solution {
    int count = 0;
    int k;
    HashMap<Integer, Integer> h = new HashMap();
    
    public void preorder(TreeNode node, int currSum) {
        if (node == null)
            return;
        currSum += node.val;
        if (currSum == k)
            count++;
        count += h.getOrDefault(currSum - k, 0);
        h.put(currSum, h.getOrDefault(currSum, 0) + 1);
        preorder(node.left, currSum);
        preorder(node.right, currSum);

        // remove the current sum from the hashmap
        // in order not to use it during 
        // the parallel subtree processing
        h.put(currSum, h.get(currSum) - 1);
    }    
    public int pathSum(TreeNode root, int sum) {
        k = sum;
        preorder(root, 0);
        return count;
    }
}