LeetCode #2096 — MEDIUM

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

Move from brute-force thinking to an efficient approach using tree strategy.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[5,1,2,3,null,6,4]
3
6

Example 2

[2,1]
2
1

Core Insight

What unlocks the optimal approach

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.
Find the path strings from root → s, and root → t. Can you use these two strings to prepare the final answer?
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'.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2096: Step-By-Step Directions From a Binary Tree Node to Another
class Solution {
    public String getDirections(TreeNode root, int startValue, int destValue) {
        TreeNode node = lca(root, startValue, destValue);
        StringBuilder pathToStart = new StringBuilder();
        StringBuilder pathToDest = new StringBuilder();
        dfs(node, startValue, pathToStart);
        dfs(node, destValue, pathToDest);
        return "U".repeat(pathToStart.length()) + pathToDest.toString();
    }

    private TreeNode lca(TreeNode node, int p, int q) {
        if (node == null || node.val == p || node.val == q) {
            return node;
        }
        TreeNode left = lca(node.left, p, q);
        TreeNode right = lca(node.right, p, q);
        if (left != null && right != null) {
            return node;
        }
        return left != null ? left : right;
    }

    private boolean dfs(TreeNode node, int x, StringBuilder path) {
        if (node == null) {
            return false;
        }
        if (node.val == x) {
            return true;
        }
        path.append('L');
        if (dfs(node.left, x, path)) {
            return true;
        }
        path.setCharAt(path.length() - 1, 'R');
        if (dfs(node.right, x, path)) {
            return true;
        }
        path.deleteCharAt(path.length() - 1);
        return false;
    }
}

// Accepted solution for LeetCode #2096: Step-By-Step Directions From a Binary Tree Node to Another
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func getDirections(root *TreeNode, startValue int, destValue int) string {
	var lca func(node *TreeNode, p, q int) *TreeNode
	lca = func(node *TreeNode, p, q int) *TreeNode {
		if node == nil || node.Val == p || node.Val == q {
			return node
		}
		left := lca(node.Left, p, q)
		right := lca(node.Right, p, q)
		if left != nil && right != nil {
			return node
		}
		if left != nil {
			return left
		}
		return right
	}
	var dfs func(node *TreeNode, x int, path *[]byte) bool
	dfs = func(node *TreeNode, x int, path *[]byte) bool {
		if node == nil {
			return false
		}
		if node.Val == x {
			return true
		}
		*path = append(*path, 'L')
		if dfs(node.Left, x, path) {
			return true
		}
		(*path)[len(*path)-1] = 'R'
		if dfs(node.Right, x, path) {
			return true
		}
		*path = (*path)[:len(*path)-1]
		return false
	}

	node := lca(root, startValue, destValue)
	pathToStart := []byte{}
	pathToDest := []byte{}
	dfs(node, startValue, &pathToStart)
	dfs(node, destValue, &pathToDest)
	return string(bytes.Repeat([]byte{'U'}, len(pathToStart))) + string(pathToDest)
}

# Accepted solution for LeetCode #2096: Step-By-Step Directions From a Binary Tree Node to Another
# 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:
        def lca(node: Optional[TreeNode], p: int, q: int):
            if node is None or node.val in (p, q):
                return node
            left = lca(node.left, p, q)
            right = lca(node.right, p, q)
            if left and right:
                return node
            return left or right

        def dfs(node: Optional[TreeNode], x: int, path: List[str]):
            if node is None:
                return False
            if node.val == x:
                return True
            path.append("L")
            if dfs(node.left, x, path):
                return True
            path[-1] = "R"
            if dfs(node.right, x, path):
                return True
            path.pop()
            return False

        node = lca(root, startValue, destValue)

        path_to_start = []
        path_to_dest = []

        dfs(node, startValue, path_to_start)
        dfs(node, destValue, path_to_dest)

        return "U" * len(path_to_start) + "".join(path_to_dest)

// Accepted solution for LeetCode #2096: Step-By-Step Directions From a Binary Tree Node to Another
/**
 * [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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/11/15/eg1.png" style="width: 214px; height: 163px;" />
 * Input: root = [5,1,2,3,null,6,4], startValue = 3, destValue = 6
 * Output: "UURL"
 * Explanation: The shortest path is: 3 &rarr; 1 &rarr; 5 &rarr; 2 &rarr; 6.
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/11/15/eg2.png" style="width: 74px; height: 102px;" />
 * Input: root = [2,1], startValue = 2, destValue = 1
 * Output: "L"
 * Explanation: The shortest path is: 2 &rarr; 1.
 *
 *  
 * Constraints:
 *
 * 	The number of nodes in the tree is n.
 * 	2 <= n <= 10^5
 * 	1 <= Node.val <= n
 * 	All the values in the tree are unique.
 * 	1 <= startValue, destValue <= n
 * 	startValue != destValue
 *
 */
pub struct Solution {}
use crate::util::tree::{TreeNode, to_tree};

// problem: https://leetcode.com/problems/step-by-step-directions-from-a-binary-tree-node-to-another/
// discuss: https://leetcode.com/problems/step-by-step-directions-from-a-binary-tree-node-to-another/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn get_directions(
        root: Option<Rc<RefCell<TreeNode>>>,
        start_value: i32,
        dest_value: i32,
    ) -> String {
        String::new()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2096_example_1() {
        let root = tree![5, 1, 2, 3, null, 6, 4];
        let start_value = 3;
        let dest_value = 6;

        let result = "UURL".to_string();

        assert_eq!(
            Solution::get_directions(root, start_value, dest_value),
            result
        );
    }

    #[test]
    #[ignore]
    fn test_2096_example_2() {
        let root = tree![2, 1];
        let start_value = 2;
        let dest_value = 1;

        let result = "L".to_string();

        assert_eq!(
            Solution::get_directions(root, start_value, dest_value),
            result
        );
    }
}

// Accepted solution for LeetCode #2096: Step-By-Step Directions From a Binary Tree Node to Another
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function getDirections(root: TreeNode | null, startValue: number, destValue: number): string {
    const lca = (node: TreeNode | null, p: number, q: number): TreeNode | null => {
        if (node === null || [p, q].includes(node.val)) {
            return node;
        }
        const left = lca(node.left, p, q);
        const right = lca(node.right, p, q);

        return left && right ? node : (left ?? right);
    };

    const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {
        if (node === null) {
            return false;
        }
        if (node.val === x) {
            return true;
        }
        path.push('L');
        if (dfs(node.left, x, path)) {
            return true;
        }
        path[path.length - 1] = 'R';
        if (dfs(node.right, x, path)) {
            return true;
        }
        path.pop();
        return false;
    };

    const node = lca(root, startValue, destValue);
    const pathToStart: string[] = [];
    const pathToDest: string[] = [];
    dfs(node, startValue, pathToStart);
    dfs(node, destValue, pathToDest);
    return 'U'.repeat(pathToStart.length) + pathToDest.join('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.