标签：time eve src vat this end short nec ogr
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 and 1000.0 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 }
[LeetCode 988] Smallest String Starting From Leaf
标签：time eve src vat this end short nec ogr
原文地址：https://www.cnblogs.com/lz87/p/10352592.html
