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
/**
* 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 {
public int pathSum(TreeNode root, long targetSum) {
if (root == null) {
return 0;
}
int ret = rootSum(root, targetSum);
ret += pathSum(root.left, targetSum);
ret += pathSum(root.right, targetSum);
return ret;
}
public int rootSum(TreeNode root, long targetSum) {
int ret = 0;
if (root == null) {
return 0;
}
int val = root.val;
if (val == targetSum) {
ret++;
}
ret += rootSum(root.left, targetSum - val);
ret += rootSum(root.right, targetSum - val);
return ret;
}
}