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Problem: 988. 从叶结点开始的最小字符串

题目描述

思路

遍历思想(利用二叉树的先序遍历)

在先序遍历的过程中，用一个变量path拼接记录下其组成的字符串，当遇到根节点时再将其反转并比较大小（字典顺序大小）同时更新较小的结果字符串（其中要注意的操作是，字符串的反转与更新）

复杂度

时间复杂度:

$O(n)$;其中$n$为二叉树的节点个数

空间复杂度:

$O(h)$；其中$h$为二叉树的高度

Code

/*** 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 String smallestFromLeaf(TreeNode root) {traverse(root);return res;}StringBuilder path = new StringBuilder();String res = null;private void traverse(TreeNode root) {if (root == null) {return;}if (root.left == null && root.right == null) {// Find the leaf node and compare the path// with the smallest lexicographic orderpath.append((char)('a' + root.val));// The resulting string is from leaf to root,// so inversion is requiredpath.reverse();String s = path.toString();if (res == null || res.compareTo(s) > 0) {res = s;}// Recover and properly maintain elements in pathpath.reverse();path.deleteCharAt(path.length() - 1);return;}path.append((char)('a' + root.val));traverse(root.left);traverse(root.right);path.deleteCharAt(path.length() - 1);}
}