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

Use recursion and traverse the tree and get number of paths with current root. Now recursively call pathSum(root.left, sum) and pathSum(root.right, sum) to get all result from left and right children.

public int pathSum(TreeNode root, int sum) {
        if (root == null) {
            return 0;
        }
        return getPathSum(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }
    
    private int getPathSum(TreeNode root, int sum) {
        int rst = 0;
        if (root == null) {
            return rst;
        }
        if (sum == root.val) {
            rst++;
        }
        rst += getPathSum(root.left, sum - root.val);
        rst += getPathSum(root.right, sum - root.val);
        return rst;
    }