# Path Sum III

Given the root of a binary tree and an integer targetSum, return *the number of paths where the sum of the values along the path equals* targetSum.

The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).

```
public int pathSum(TreeNode root, int sum) {
HashMap<Integer, Integer> preSum = new HashMap();
preSum.put(0,1);
return helper(root, 0, sum, preSum);
}
public int helper(TreeNode root, int currSum, int target, HashMap<Integer, Integer> preSum) {
if (root == null) {
return 0;
}
currSum += root.val;
int res = preSum.getOrDefault(currSum - target, 0);
preSum.put(currSum, preSum.getOrDefault(currSum, 0) + 1);
res += helper(root.left, currSum, target, preSum) + helper(root.right, currSum, target, preSum);
preSum.put(currSum, preSum.get(currSum) - 1);
return res;
}
```

## Related Problems

Given the head of a linked list, remove the nth node from the end of the list and return its head.

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one **sorted** list. The list should be made by splicing together the nodes of the first two lists.

Return *the head of the merged linked list*.

Given the root of a binary tree, flatten the tree into a "linked list":

The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.

The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Given the head of a singly linked list, return true* if it is a palindrome or *false* otherwise*.