package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

/**
 * 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
 */

public class _437 {

    public int pathSum(TreeNode root, int sum) {
        if (root == null) {
            return 0;
        }
        return pathSumFrom(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }

    private int pathSumFrom(TreeNode root, int sum) {
        if (root == null) {
            return 0;
        }
        return (root.val == sum ? 1 : 0) + pathSumFrom(root.left, sum - root.val) + pathSumFrom(root.right, sum - root.val);
    }

}

package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

/**
 * 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
 */

public class _437 {

    public static class Solution1 {
        public int pathSum(TreeNode root, int sum) {
            if (root == null) {
                return 0;
            }
            return pathSumFrom(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
        }

        private int pathSumFrom(TreeNode root, int sum) {
            if (root == null) {
                return 0;
            }
            return (root.val == sum ? 1 : 0) + pathSumFrom(root.left, sum - root.val) + pathSumFrom(root.right, sum - root.val);
        }
    }

}

refactor 437