988. Smallest String Starting From Leaf

Description

You are given the root of a binary tree where each node has a value in the range [0, 25] representing the letters 'a' to 'z'.

Return the lexicographically smallest string that starts at a leaf of this tree and ends at the root.

As a reminder, any shorter prefix of a string is lexicographically smaller.

• For example, "ab" is lexicographically smaller than "aba".

A leaf of a node is a node that has no children.

Example 1:

Input: root = [0,1,2,3,4,3,4]
Output: "dba"


Example 2:

Input: root = [25,1,3,1,3,0,2]


Example 3:

Input: root = [2,2,1,null,1,0,null,0]
Output: "abc"


Constraints:

• The number of nodes in the tree is in the range [1, 8500].
• 0 <= Node.val <= 25

Solutions

• /**
* 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.
# 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.
* 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
}