257. Binary Tree Paths - Easy

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

 

backtracking, 记得返回上层的时候删掉该层节点

time: O(n) 　　-- visited each node exactly once, space: O(logn)　　-- worst case: O(n) unbalanced tree

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        dfs(root, new StringBuilder(), res);
        return res;
    }
    
    public void dfs(TreeNode root, StringBuilder sb, List<String> res) {
        if(root == null) {
            return;
        }
        int len = sb.length();
        sb.append(root.val);
        if(root.left == null && root.right == null) {
            res.add(sb.toString());
            sb.delete(len, sb.length());
            return;
        }
        
        sb.append("->");
        
        dfs(root.left, sb, res);
        dfs(root.right, sb, res);
        sb.delete(len, sb.length());
    }
}