94 lines
2.2 KiB
Java
94 lines
2.2 KiB
Java
//给你二叉树的根节点 root 和一个表示目标和的整数 targetSum ，判断该树中是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和
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// targetSum 。
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//
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// 叶子节点 是指没有子节点的节点。
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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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//输入：root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
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//输出：true
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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 = [1,2,3], targetSum = 5
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//输出：false
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//
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//
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// 示例 3：
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//
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//
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//输入：root = [1,2], targetSum = 0
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//输出：false
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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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// 树中节点的数目在范围 [0, 5000] 内
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// -1000 <= Node.val <= 1000
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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 👍 702 👎 0
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package leetcode.editor.cn;
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import com.code.leet.entiy.TreeNode;
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//112:路径总和
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class PathSum {
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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 PathSum().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 boolean hasPathSum(TreeNode root, int targetSum) {
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if (root == null) {
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return false;
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}
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return dfs(root, 0, targetSum);
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}
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private boolean dfs(TreeNode root, int sum, int targetSum) {
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if (root.left == null && root.right == null) {
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return root.val + sum == targetSum;
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}
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if (root.left != null) {
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if (dfs(root.left, sum + root.val, targetSum)) {
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return true;
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}
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}
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if (root.right != null) {
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return dfs(root.right, sum + root.val, targetSum);
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}
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return false;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |