题目
:::info
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”]
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/binary-tree-paths
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
:::
代码
/**
* 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 List<String> binaryTreePaths(TreeNode root) {
List<String> paths = new ArrayList<String>();
dfs(root,"",paths);
return paths;
}
//path为当前正在遍历的路径,paths为最终的路径集合
public static void dfs(TreeNode root,String path,List<String> paths){
if(root != null){
StringBuffer pathSB = new StringBuffer(path);
pathSB.append(Integer.toString(root.val));
if(root.left == null && root.right == null){ //节点为叶子节点
paths.add(pathSB.toString()); //集合中存放的是字符串
}else{ //节点为非叶子节点
pathSB.append("->");
dfs(root.left,pathSB.toString(),paths);
dfs(root.right,pathSB.toString(),paths);
}
}
}
}