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

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

Hint:

1. The shortest path between any two nodes in a tree must pass through their Lowest Common Ancestor (LCA). The path will travel upwards from node s to the LCA and then downwards from the LCA to node t.
2. Find the path strings from root → s, and root → t. Can you use these two strings to prepare the final answer?
3. Remove the longest common prefix of the two path strings to get the path LCA → s, and LCA → t. Each step in the path of LCA → s should be reversed as 'U'.

Solution:

To solve this problem, we can follow these steps:

1. Find the path from the root to the start node (s) and the destination node (t): This can be done using Depth-First Search (DFS).
2. Find the Lowest Common Ancestor (LCA) of the start and destination nodes: The LCA is the lowest node in the tree that has both s and t as descendants.
3. Construct the path from s to the LCA and from the LCA to t: The path from s to the LCA will be in reverse order and all directions will be 'U'. The path from the LCA to t will be in the natural order.
4. Combine these paths to form the final path: This will give the shortest path from s to t.

Let's implement this solution in PHP: 2096. Step-By-Step Directions From a Binary Tree Node to Another

<?php
// Example usage:
$root = new TreeNode(5);
$root->left = new TreeNode(1);
$root->right = new TreeNode(2);
$root->left->left = new TreeNode(3);
$root->right->left = new TreeNode(6);
$root->right->right = new TreeNode(4);

$startValue = 3;
$destValue = 6;
echo getDirections($root, $startValue, $destValue); // Outputs: "UURL"

$root2 = new TreeNode(5);
$root2->left = new TreeNode(1);
$root2->right = new TreeNode(2);
$root2->left->left = new TreeNode(3);
$root2->right->left = new TreeNode(6);
$root2->right->right = new TreeNode(4);

$startValue2 = 3;
$destValue2 = 6;
echo getDirections($root2, $startValue2, $destValue2); // Outputs: "L"

?>

Explanation:

1. TreeNode Class: A simple class to represent a node in the binary tree.
2. findPath Function: This function finds the path from the root to the given value and stores the path in the provided array ($path). It uses DFS and marks each step with 'L' for left and 'R' for right.
3. getDirections Function: This function uses findPath to get the paths from the root to startValue and destValue. It then finds the common path length (LCA). The number of 'U' moves is the difference in length between the start path and the common path. The remaining path to the destination node is appended.
4. Example Usage: Creates the binary tree as shown in the example, and calls getDirections with startValue and destValue. The result is the string representing the shortest path.

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