Question

988. 从叶结点开始的最小字符串

English Version

题目描述

给定一颗根结点为 root 的二叉树，树中的每一个结点都有一个 [0, 25] 范围内的值，分别代表字母 'a' 到 'z'。

返回 _按字典序最小 的字符串，该字符串从这棵树的一个叶结点开始，到根结点结束_。

注：字符串中任何较短的前缀在 字典序上 都是 较小 的：

• 例如，在字典序上 "ab" 比 "aba" 要小。叶结点是指没有子结点的结点。 

节点的叶节点是没有子节点的节点。

 

示例 1：

输入：root = [0,1,2,3,4,3,4]
输出："dba"

示例 2：

输入：root = [25,1,3,1,3,0,2]
输出："adz"

示例 3：

输入：root = [2,2,1,null,1,0,null,0]
输出："abc"

 

提示：

• 给定树的结点数在 [1, 8500] 范围内
• 0 <= Node.val <= 25

解法

方法一

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def smallestFromLeaf(self, root: TreeNode) -> str:
    ans = chr(ord('z') + 1)

    def dfs(root, path):
      nonlocal ans
      if root:
        path.append(chr(ord('a') + root.val))
        if root.left is None and root.right is None:
          ans = min(ans, ''.join(reversed(path)))
        dfs(root.left, path)
        dfs(root.right, path)
        path.pop()

    dfs(root, [])
    return ans

/**
 * 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 {
  private StringBuilder path;
  private String ans;

  public String smallestFromLeaf(TreeNode root) {
    path = new StringBuilder();
    ans = String.valueOf((char) ('z' + 1));
    dfs(root, path);
    return ans;
  }

  private void dfs(TreeNode root, StringBuilder path) {
    if (root != null) {
      path.append((char) ('a' + root.val));
      if (root.left == null && root.right == null) {
        String t = path.reverse().toString();
        if (t.compareTo(ans) < 0) {
          ans = t;
        }
        path.reverse();
      }
      dfs(root.left, path);
      dfs(root.right, path);
      path.deleteCharAt(path.length() - 1);
    }
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  string ans = "";

  string smallestFromLeaf(TreeNode* root) {
    string path = "";
    dfs(root, path);
    return ans;
  }

  void dfs(TreeNode* root, string& path) {
    if (!root) return;
    path += 'a' + root->val;
    if (!root->left && !root->right) {
      string t = path;
      reverse(t.begin(), t.end());
      if (ans == "" || t < ans) ans = t;
    }
    dfs(root->left, path);
    dfs(root->right, path);
    path.pop_back();
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func smallestFromLeaf(root *TreeNode) string {
  ans := ""
  var dfs func(root *TreeNode, path string)
  dfs = func(root *TreeNode, path string) {
    if root == nil {
      return
    }
    path = string('a'+root.Val) + path
    if root.Left == nil && root.Right == nil {
      if ans == "" || path < ans {
        ans = path
      }
      return
    }
    dfs(root.Left, path)
    dfs(root.Right, path)
  }

  dfs(root, "")
  return ans
}