437. Path Sum III

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

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

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

// Brute Force
int helper(TreeNode* node, int sum) {
    if (!node) return 0;
    return (node->val == sum ? 1 : 0) + helper(node->left, sum - node->val) + helper(node->right, sum - node->val);
}
int pathSum(TreeNode* root, int sum) { // time: O(n^2); space: O(n)
    if (!root) return 0;
    return helper(root, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}

// Use hashmap to avoid duplicate calculation
int helper(TreeNode* node, int sum, int curSum, unordered_map<int, int>& m) { // time: O(n); space: O(n)
    if (!node) return 0;
    curSum += node->val;
    int res = m[curSum - sum];
    ++m[curSum];
    res += helper(node->left, sum, curSum, m) + helper(node->right, sum, curSum, m);
    --m[curSum];
    return res;
}
int pathSum(TreeNode* root, int sum) {
    unordered_map<int, int> m;
    m[0] = 1;
    return helper(root, sum, 0, m);
}