[LeetCode 988] Smallest String Starting From Leaf

Given the root of a binary tree, each node has a value from 0 to 25 representing the letters 'a' to 'z': a value of 0 represents 'a', a value of 1 represents 'b', and so on.

Find the lexicographically smallest string that starts at a leaf of this tree and ends at the root.

(As a reminder, any shorter prefix of a string is lexicographically smaller: for example, "ab" is lexicographically smaller than "aba".  A leaf of a node is a node that has no children.)

Example 1:

Input: [0,1,2,3,4,3,4]
Output: "dba"

Example 2:

Input: [25,1,3,1,3,0,2]
Output: "adz"

Example 3:

Input: [2,2,1,null,1,0,null,0]
Output: "abc"

Note:

1. The number of nodes in the given tree will be between 1 and 1000.
2. Each node in the tree will have a value between 0 and 25.

We can use dfs to get all strings, sort them then get the smallest one. Even better, we can just keep a global variable of the current smallest string, each time we get a new string, compare it with the current smallest string and update it if necessary. This way it avoids keeping all the strings in a list and sorting them.

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     private String ans = null;
12     
13     public String smallestFromLeaf(TreeNode root) {
14         dfs(root, new StringBuilder());
15         return ans;
16     }
17     
18     private void dfs(TreeNode node, StringBuilder sb) {
19         if(node == null) {
20             return;
21         }
22         sb.append((char)('a' + node.val));
23         if(node.left == null && node.right == null) {
24             sb.reverse();
25             String s = sb.toString();
26             sb.reverse();
27             
28             if(ans == null || s.compareTo(ans) < 0) {
29                 ans = s;
30             }
31         }
32         dfs(node.left, sb);
33         dfs(node.right, sb);
34         sb.deleteCharAt(sb.length() - 1);
35     }
36 }