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
}