offer2-50

给定一个二叉树的根节点 root ，和一个整数 targetSum ，求该二叉树里节点值之和等于 targetSum 的 路径 的数目。

路径 不需要从根节点开始，也不需要在叶子节点结束，但是路径方向必须是向下的（只能从父节点到子节点）。

示例 1：

输入：root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
输出：3
解释：和等于 8 的路径有 3 条，如图所示。

示例 2：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：3

提示:

• 二叉树的节点个数的范围是 [0,1000]
• -109 <= Node.val <= 109
• -1000 <= targetSum <= 1000

//leetcode submit region begin(Prohibit modification and deletion)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int num;

    public int pathSum(TreeNode root, int targetSum) {
        this.num = 0;
        if (root == null) {
            return 0;
        }

        visit(root.val, targetSum, root, true);
        return this.num;
    }

    private void visit(long curSum, int targetSum, TreeNode node, boolean isFromRoot) {
        if (curSum == targetSum) {
            num++;
        }

        if (node.left != null) {
            visit(curSum + node.left.val, targetSum, node.left, isFromRoot);

            if (isFromRoot) {
                visit(node.left.val, targetSum, node.left, false);
            }
        }
        if (node.right != null) {
            visit(curSum + node.right.val, targetSum, node.right, isFromRoot);

            if (isFromRoot) {
                visit(node.right.val, targetSum, node.right, false);
            }
        }
    }
}
//leetcode submit region end(Prohibit modification and deletion)