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

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

Solutions (Python)

1. Solution

# 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: Optional[TreeNode], startValue: int, destValue: int) -> str:
        root2Start = self.root2(root, startValue)
        root2Dest = self.root2(root, destValue)
        minPathLen = min(len(root2Start), len(root2Dest))

        for i in range(minPathLen + 1):
            if i == minPathLen or root2Start[i] != root2Dest[i]:
                return 'U' * (len(root2Start) - i) + root2Dest[i:]

    def root2(self, root: Optional[TreeNode], destValue: int) -> Optional[str]:
        if root is None:
            return None

        if root.val == destValue:
            return ""

        root2Left = self.root2(root.left, destValue)
        if root2Left is not None:
            return 'L' + root2Left

        root2Right = self.root2(root.right, destValue)
        if root2Right is not None:
            return 'R' + root2Right

        return None