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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]
Output: "adz"

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
}

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