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
本题不能用这种以下这种方式实现:
class Solution:
def smallestFromLeaf(self, root: TreeNode) -> str:
if root is None:
return ''
left = self.smallestFromLeaf(root.left)
right = self.smallestFromLeaf(root.right)
val = chr(ord('a') + root.val)
return min(left + val, right + val)
我们举个例子来说明,对于上面这棵二叉树,正确答案应该是 "ababz",但是我们采用以上实现方式得到的答案是 "abz"。
问题就在于,当 str(x) < str(y)
,并不能保证 str(x) + a < str(y) + a
,例如 "ab" < "abab"
,但是 "abz" > "ababz"
。
本题可以用 DFS 解决,每次到达一个叶子节点时,翻转此路径上的字符串,并与 ans 比较大小,取二者较小值。
# 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
}