Problem Description

You have a binary tree where each node has a unique value from 1 to n. Given two values - startValue (the starting node) and destValue (the destination node) - you need to find the shortest path between these two nodes.

The path should be represented as a string using three types of moves:

• 'L' - move from current node to its left child
• 'R' - move from current node to its right child
• 'U' - move from current node to its parent

The solution works by first finding the Lowest Common Ancestor (LCA) of the start and destination nodes. The LCA is the deepest node that is an ancestor of both nodes.

Once we have the LCA, we find two paths:

1. Path from LCA to startValue
2. Path from LCA to destValue

To get from startValue to destValue, we need to:

1. First go UP from startValue to the LCA (this requires as many 'U' moves as the length of the path from LCA to start)
2. Then go DOWN from LCA to destValue (following the actual path with 'L' and 'R' moves)

For example, if the path from LCA to start is ['L', 'R'] and the path from LCA to destination is ['R', 'L'], the final answer would be "UU" + "RL" = "UURL" - go up twice to reach the LCA, then right and left to reach the destination.

The lca function finds the lowest common ancestor using recursion. The dfs function finds the path from a given node to a target value by exploring left and right subtrees, backtracking when necessary.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: A binary tree is a special type of graph where each node has at most two children and there are no cycles.

Is it a tree?

• Yes: The problem explicitly states we're working with a binary tree structure.

DFS

• We arrive at DFS: Since we're dealing with a tree structure, the flowchart directs us to use Depth-First Search.

Conclusion: The flowchart suggests using DFS (Depth-First Search) for this problem.

This aligns perfectly with the solution approach, which uses DFS in two key ways:

1. Finding the Lowest Common Ancestor (LCA): The lca function uses DFS to recursively traverse the tree and find the deepest node that is an ancestor of both the start and destination nodes.
2. Finding paths from LCA to target nodes: The dfs function uses DFS with backtracking to find the path from the LCA to both the start and destination values, trying left and right subtrees and backtracking when necessary.

The DFS pattern is ideal here because:

• We need to explore the tree structure deeply to find specific nodes
• We need to track paths through the tree (which DFS naturally does via recursion)
• The problem requires traversing from one node to another through parent-child relationships, which is a classic tree traversal scenario

Intuition

The key insight is that to find the shortest path between any two nodes in a tree, we must go through their Lowest Common Ancestor (LCA). Why? Because in a tree, there's only one path between any two nodes, and that path must pass through their common ancestor.

Think of it like finding directions between two locations in a family tree. If you want to go from one cousin to another, you'd first need to trace back up to their common grandparent, then trace down to the other cousin.

Here's how we can visualize the journey:

1. From the start node, we need to go UP to reach the LCA
2. From the LCA, we need to go DOWN to reach the destination

The challenge is that we can only move using 'L', 'R', and 'U' commands. So our strategy becomes:

• First, find the LCA of both nodes
• Find the path from LCA to the start node (this tells us how many steps UP we need)
• Find the path from LCA to the destination node (this gives us the exact 'L' and 'R' moves needed)

For the path from start to LCA, we don't actually need the exact moves - we just need to know the length. Since we're going backwards (from child to ancestor), every move is an 'U'. If the path from LCA to start has length 3, we need "UUU" to go back up.

For the path from LCA to destination, we need the exact sequence of 'L' and 'R' moves to navigate down the tree.

The beauty of this approach is that it guarantees the shortest path because:

• There's only one path between any two nodes in a tree
• Going through the LCA is the only way to connect them
• We're not taking any unnecessary detours

Learn more about Tree, Depth-First Search and Binary Tree patterns.

Solution Approach

The solution implements the strategy we identified using two main functions:

1. Finding the Lowest Common Ancestor (LCA)

The lca function uses a recursive DFS approach:

• Base case: If the current node is None or matches either startValue or destValue, return the current node
• Recursively search left and right subtrees
• If both left and right subtrees return non-null values, the current node is the LCA (both values are in different subtrees)
• Otherwise, return whichever subtree contains the value(s)

def lca(node, p, q):
    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

2. Finding the Path from a Node to a Target

The dfs function finds the path from a given node to a target value using backtracking:

• If we find the target, return True
• Try going left by appending 'L' to the path
• If left doesn't work, replace the last character with 'R' and try right
• If neither works, pop the character and backtrack

def dfs(node, x, path):
    if node.val == x:
        return True
    path.append("L")
    if dfs(node.left, x, path):
        return True
    path[-1] = "R"  # Replace 'L' with 'R'
    if dfs(node.right, x, path):
        return True
    path.pop()  # Backtrack
    return False

3. Combining the Results

Once we have:

• The LCA node
• Path from LCA to startValue (e.g., ['L', 'R'])
• Path from LCA to destValue (e.g., ['R', 'L'])

We construct the final answer:

• Convert the length of path_to_start into that many 'U' moves (going up from start to LCA)
• Append the exact path from LCA to destination

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

The time complexity is O(n) where n is the number of nodes, as we potentially visit each node a constant number of times. The space complexity is O(h) where h is the height of the tree, due to the recursion stack and path storage.

Example Walkthrough

Let's walk through a concrete example to see how the solution works.

Consider this binary tree:

        5
       / \
      1   2
     /   / \
    3   6   4

We want to find the path from startValue = 3 to destValue = 6.

Step 1: Find the Lowest Common Ancestor (LCA)

Starting from root (5), we search for nodes 3 and 6:

• At node 5: Check left subtree (contains 3) and right subtree (contains 6)
• Left subtree returns node 1 (ancestor of 3)
• Right subtree returns node 2 (ancestor of 6)
• Since both subtrees return non-null, node 5 is the LCA

Step 2: Find Path from LCA to Start (5 → 3)

Starting from LCA (5), find path to node 3:

• Try left: Go to node 1 (append 'L')
• From node 1, try left: Go to node 3 (append 'L')
• Found target! Path = ['L', 'L']

Step 3: Find Path from LCA to Destination (5 → 6)

Starting from LCA (5), find path to node 6:

• Try left: Go to node 1 (append 'L')
• Node 1 has only left child (3), not our target
• Backtrack, replace 'L' with 'R': Go to node 2
• From node 2, try left: Go to node 6 (append 'L')
• Found target! Path = ['R', 'L']

Step 4: Construct Final Answer

• Path from LCA to start: ['L', 'L'] → Length is 2, so we need "UU" to go up
• Path from LCA to destination: ['R', 'L'] → Use "RL" to go down
• Final answer: "UU" + "RL" = "UURL"

This means: From node 3, go up twice to reach node 5 (the LCA), then go right to node 2, then left to reach node 6.

Let's verify: 3 → (U) → 1 → (U) → 5 → (R) → 2 → (L) → 6 ✓

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm consists of three main operations:

1. Finding the Lowest Common Ancestor (LCA): The lca function visits each node at most once in the worst case, resulting in O(n) time complexity.
2. Finding path from LCA to startValue: The dfs function traverses the tree to find the target node, visiting each node at most once, which takes O(n) time in the worst case.
3. Finding path from LCA to destValue: Similarly, this also takes O(n) time in the worst case.

Since these operations are performed sequentially, the overall time complexity is O(n) + O(n) + O(n) = O(n), where n is the number of nodes in the binary tree.

Space Complexity: O(n)

The space complexity comes from:

1. Recursion stack space: Both lca and dfs functions use recursion. In the worst case (skewed tree), the recursion depth can be O(n).
2. Path storage: The path_to_start and path_to_dest lists store the paths from LCA to the respective nodes. In the worst case, these paths can have length O(n).
3. The final result string construction also takes O(n) space in the worst case.

Therefore, the overall space complexity is O(n), where n is the number of nodes in the binary tree.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Handling the Case When Start or Destination is the LCA

One common pitfall is not properly handling the edge case where either the startValue or destValue is itself the LCA. When the LCA function returns early because it finds one of the target nodes, that node might actually be the ancestor of the other node.

Problem Example: Consider a tree where node 5 is the parent of node 3:

    5 (startValue)
   /
  3 (destValue)

The LCA function will return node 5 immediately upon finding it, which is correct. However, if we're not careful with the path finding, we might get incorrect results.

Solution: The current implementation handles this correctly by:

1. The LCA function properly returns the ancestor node even if it's one of the targets
2. The find_path function will return an empty path when searching from the LCA to itself (if start is the LCA)
3. The final result correctly uses "U" * 0 (empty string) when the path from LCA to start is empty

2. Mutating the Path List Incorrectly During Backtracking

A subtle but critical pitfall occurs in the find_path function with the line path[-1] = "R". Developers might mistakenly write:

# WRONG approach:
path.append("L")
if find_path(node.left, target, path):
    return True
path.pop()  # Remove 'L'
path.append("R")  # Add 'R'
if find_path(node.right, target, path):
    return True
path.pop()  # Remove 'R'

Why this is inefficient: This performs unnecessary list operations (pop and append) when switching from left to right exploration.

Correct approach (as in the solution):

path.append("L")
if find_path(node.left, target, path):
    return True
path[-1] = "R"  # Simply replace 'L' with 'R'
if find_path(node.right, target, path):
    return True
path.pop()  # Only pop once at the end

This optimization reduces the number of list operations and makes the code cleaner.

3. Not Handling Null Nodes in the DFS Path Finding

A common mistake is forgetting to check for null nodes in the find_path function:

Wrong:

def find_path(node, target, path):
    if node.val == target:  # This will crash if node is None!
        return True
    # ... rest of the code

Correct (as in the solution):

def find_path(node, target, path):
    if node is None:  # Check for None first
        return False
    if node.val == target:
        return True
    # ... rest of the code

4. Assuming the Tree Structure Without Validation

The solution assumes that both startValue and destValue exist in the tree and are unique. In a production environment, you should validate:

• Both values actually exist in the tree
• The tree is properly formed (no cycles, proper parent-child relationships)

Enhanced validation approach:

def getDirections(self, root, startValue, destValue):
    # First, verify both nodes exist
    def node_exists(node, value):
        if not node:
            return False
        if node.val == value:
            return True
        return node_exists(node.left, value) or node_exists(node.right, value)
  
    if not node_exists(root, startValue) or not node_exists(root, destValue):
        return ""  # Or raise an exception
  
    # Continue with the main logic...

5. String Concatenation Performance Issues

While the current solution is fine for typical tree sizes, for very deep trees with long paths, string concatenation could become a performance bottleneck:

Current approach:

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

For extremely large paths, consider using a list:

result = []
result.extend(['U'] * len(path_to_start))
result.extend(path_to_dest)
return "".join(result)

This approach is more memory-efficient for very large strings, though the difference is negligible for typical problem constraints.