package tree;

import java.util.HashMap;
import java.util.Map;

/**
 * Created by gouthamvidyapradhan on 08/04/2017.
 * You are given a binary tree in which each node contains an integer value.
 * <p>
 * Find the number of paths that sum to a given value.
 * <p>
 * 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).
 * <p>
 * The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
 * <p>
 * Example:
 * <p>
 * root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
 * <p>
 * 10
 * /  \
 * 5   -3
 * / \    \
 * 3   2   11
 * / \   \
 * 3  -2   1
 * <p>
 * Return 3. The paths that sum to 8 are:
 * <p>
 * 1.  5 -> 3
 * 2.  5 -> 2 -> 1
 * 3. -3 -> 11
 */

public class PathSumIII {
    /**
     *
     */
    public static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }

    private Map<Integer, Integer> pathCount = new HashMap<>();
    private int totalCount;

    public static void main(String[] args) throws Exception {
        TreeNode node = new TreeNode(1);
        System.out.println(new PathSumIII().pathSum(node, 0));
    }

    public int pathSum(TreeNode root, int sum) {
        if (root == null) return 0;
        dfs(root, sum, 0);
        return totalCount;
    }

    private void dfs(TreeNode root, int target, int pSum) {
        if (root != null) {
            pSum += root.val;
            if (pSum == target) totalCount++;
            totalCount += pathCount.getOrDefault(pSum - target, 0);
            pathCount.put(pSum, pathCount.getOrDefault(pSum, 0) + 1);
            dfs(root.left, target, pSum);
            dfs(root.right, target, pSum);
            pathCount.put(pSum, pathCount.get(pSum) - 1);
        }
    }
}