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//给你一个二叉树的根节点 root ,按 任意顺序 ,返回所有从根节点到叶子节点的路径。
//
// 叶子节点 是指没有子节点的节点。
//
// 示例 1:
//
//
//输入:root = [1,2,3,null,5]
//输出:["1->2->5","1->3"]
//
//
// 示例 2:
//
//
//输入:root = [1]
//输出:["1"]
//
//
//
//
// 提示:
//
//
// 树中节点的数目在范围 [1, 100] 内
// -100 <= Node.val <= 100
//
//
// Related Topics 树 深度优先搜索 字符串 回溯 二叉树 👍 1072 👎 0//leetcode submit region begin(Prohibit modification and deletion)import java.util.ArrayList;/*** 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 {/*** 记录结果*/List<String> res = new ArrayList<>();public List<String> binaryTreePaths(TreeNode root) {if(root == null){return res;}findWay(new ArrayList<>(), root);return res;}private void findWay(List<Integer> recordList, TreeNode cur) {recordList.add(cur.val);if (cur.left == null && cur.right == null) {StringBuilder s = new StringBuilder();for (int i = 0; i < recordList.size() - 1; i++) {s.append(recordList.get(i)).append("->");}s.append(recordList.get(recordList.size() - 1));res.add(s.toString());return;}if (cur.left != null) {findWay(recordList, cur.left);recordList.remove(recordList.size() - 1);}if (cur.right != null) {findWay(recordList, cur.right);recordList.remove(recordList.size() - 1);}return;}
}
//leetcode submit region end(Prohibit modification and deletion)
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