91 lines
2.3 KiB
Java
91 lines
2.3 KiB
Java
//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。
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//
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// 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）。
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//
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//
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//
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// 示例 1：
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//
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//
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//
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//
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//输入：root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
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//输出：3
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//解释：和等于 8 的路径有 3 条，如图所示。
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//
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//
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// 示例 2：
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//
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//
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//输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
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//输出：3
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//
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//
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//
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//
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// 提示:
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//
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//
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// 二叉树的节点个数的范围是 [0,1000]
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// -10⁹ <= Node.val <= 10⁹
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// -1000 <= targetSum <= 1000
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//
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// Related Topics 树 深度优先搜索 二叉树 👍 1091 👎 0
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package leetcode.editor.cn;
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import com.code.leet.entiy.TreeNode;
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//437:路径总和 III
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class PathSumIii {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new PathSumIii().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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public int pathSum(TreeNode root, int targetSum) {
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if (root == null) {
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return 0;
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}
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int ret = rootSum(root, targetSum);
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ret += pathSum(root.left, targetSum);
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ret += pathSum(root.right, targetSum);
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return ret;
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}
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public int rootSum(TreeNode root, int targetSum) {
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int ret = 0;
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if (root == null) {
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return 0;
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}
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int val = root.val;
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if (val == targetSum) {
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ret++;
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}
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ret += rootSum(root.left, targetSum - val);
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ret += rootSum(root.right, targetSum - val);
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return ret;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |