LeetCode-in-Java

2096. Step-By-Step Directions From a Binary Tree Node to Another

Medium

You are given the root of a binary tree with n nodes. Each node is uniquely assigned a value from 1 to n. You are also given an integer startValue representing the value of the start node s, and a different integer destValue representing the value of the destination node t.

Find the shortest path starting from node s and ending at node t. Generate step-by-step directions of such path as a string consisting of only the uppercase letters 'L', 'R', and 'U'. Each letter indicates a specific direction:

• 'L' means to go from a node to its left child node.
• 'R' means to go from a node to its right child node.
• 'U' means to go from a node to its parent node.

Return the step-by-step directions of the shortest path from node s to node t.

Example 1:

Input: root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6

Output: “UURL”

Explanation: The shortest path is: 3 → 1 → 5 → 2 → 6.

Example 2:

Input: root = [2,1], startValue = 2, destValue = 1

Output: “L”

Explanation: The shortest path is: 2 → 1.

Constraints:

• The number of nodes in the tree is n.
• 2 <= n <= 105
• 1 <= Node.val <= n
• All the values in the tree are unique.
• 1 <= startValue, destValue <= n
• startValue != destValue

Solution

import com_github_leetcode.TreeNode;

public class Solution {
    private boolean find(TreeNode n, int val, StringBuilder sb) {
        if (n.val == val) {
            return true;
        }
        if (n.left != null && find(n.left, val, sb)) {
            sb.append("L");
        } else if (n.right != null && find(n.right, val, sb)) {
            sb.append("R");
        }
        return !sb.isEmpty();
    }

    public String getDirections(TreeNode root, int startValue, int destValue) {
        StringBuilder s = new StringBuilder();
        StringBuilder d = new StringBuilder();
        find(root, startValue, s);
        find(root, destValue, d);
        int i = 0;
        int maxI = Math.min(d.length(), s.length());
        while (i < maxI && s.charAt(s.length() - i - 1) == d.charAt(d.length() - i - 1)) {
            ++i;
        }
        return "U".repeat(Math.max(0, s.length() - i)) + d.reverse().substring(i);
    }
}

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