力扣988. 从叶结点开始的最小字符串

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 order
            path.append((char)('a' + root.val));
            // The resulting string is from leaf to root,
            // so inversion is required
            path.reverse();
            String s = path.toString();
            if (res == null || res.compareTo(s) > 0) {
                res = s;
            }

            // Recover and properly maintain elements in path
            path.reverse();
            path.deleteCharAt(path.length() - 1);
            return;
        }
        
        path.append((char)('a' + root.val));
        traverse(root.left);
        traverse(root.right);
        path.deleteCharAt(path.length() - 1);
    }
}