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

ou 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

# 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 getDirections(self, root, startValue, destValue):

        def lca(node):
            if not node:
                return None

            if node.val == startValue or node.val == destValue:
                return node

            left = lca(node.left)
            right = lca(node.right)

            if left and right:
                return node

            return left or right

        def findPath(node, target, path):
            if not node:
                return False

            if node.val == target:
                return True

            path.append("L")
            if findPath(node.left, target, path):
                return True
            path.pop()

            path.append("R")
            if findPath(node.right, target, path):
                return True
            path.pop()

            return False

        ancestor = lca(root)

        pathStart = []
        pathDest = []

        findPath(ancestor, startValue, pathStart)
        findPath(ancestor, destValue, pathDest)

        return "U" * len(pathStart) + "".join(pathDest)