Problem Description

You have a binary tree where each node contains a value between 0 and 25. These values represent letters of the alphabet, where 0 represents 'a', 1 represents 'b', and so on up to 25 representing 'z'.

The task is to find the lexicographically smallest string that can be formed by traveling from any leaf node up to the root node.

Key points to understand:

• A leaf node is a node that has no children (neither left nor right child)
• The string is formed by reading the node values from a leaf to the root (bottom-up), converting each value to its corresponding letter
• Lexicographical order is dictionary order - a string is smaller if it would appear earlier in a dictionary
  • For example, "ab" comes before "aba" in dictionary order, making "ab" lexicographically smaller
  • When comparing strings, we check character by character from left to right

The solution uses depth-first search (DFS) to explore all paths from the root to each leaf. During traversal:

1. It builds the path from root to leaf by appending characters
2. When reaching a leaf node, it reverses the path (since we need leaf-to-root order) and compares it with the current smallest string
3. It backtracks by removing the last character when returning from a node
4. The initial answer is set to chr(ord('z') + 1) which is a character after 'z', ensuring any valid path will be smaller

For example, if you have a tree where the root is 0 ('a'), left child is 1 ('b'), and right child is 2 ('c'), both children being leaves, the possible strings would be "ba" (from left leaf to root) and "ca" (from right leaf to root). The answer would be "ba" as it's lexicographically smaller.

Flowchart Walkthrough

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

Is it a graph?

• Yes: We have a binary tree, which is a special type of graph with nodes (tree nodes) and edges (parent-child relationships).

Is it a tree?

• Yes: The problem explicitly states we're working with a binary tree structure where we have a root node and each node can have at most two children (left and right).

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 makes perfect sense because:

1. We need to explore all possible paths from leaves to the root
2. DFS naturally traverses from root to leaves, allowing us to build paths incrementally
3. When we reach a leaf node, we can evaluate the complete path (leaf to root)
4. DFS with backtracking lets us efficiently explore all paths by maintaining the current path and removing nodes as we backtrack
5. The tree structure guarantees no cycles, making DFS a safe and efficient choice

The DFS approach in the solution:

• Recursively traverses the tree from root to leaves
• Maintains a path list to track the current path from root to leaf
• At each leaf, reverses the path to get the leaf-to-root string and compares it with the current minimum
• Uses backtracking (via path.pop()) to explore all possible paths efficiently

Intuition

The key insight is that we need to compare strings formed by traveling from different leaf nodes to the root. Since we want the lexicographically smallest string, we must examine all possible leaf-to-root paths.

Think about how we'd solve this manually: we'd identify all leaf nodes, trace each path back to the root while collecting characters, then compare all resulting strings. This naturally suggests a traversal approach where we visit every leaf.

DFS is perfect here because:

1. It naturally reaches leaf nodes by going as deep as possible
2. It maintains the current path as it traverses, which is exactly what we need to build our string
3. When backtracking, it automatically "undoes" the path, allowing us to explore other branches

The clever part is building the path as we go down (root to leaf), then reversing it when we reach a leaf. Why? Because DFS naturally traverses from root to leaf, but we need the string from leaf to root. Rather than traversing upward (which would be complex), we simply reverse the accumulated path.

Consider a small example: if we have root 'a' with left child 'b' (a leaf), our DFS would:

• Start at 'a', add it to path: ['a']
• Move to 'b', add it to path: ['a', 'b']
• Detect it's a leaf, reverse to get "ba" (leaf to root)
• Compare with our current minimum

The initialization with chr(ord('z') + 1) is a sentinel value - any valid path will be lexicographically smaller than a character beyond 'z', ensuring our first valid path becomes the initial minimum.

The backtracking step (path.pop()) is crucial - it removes the current node from the path when we're done exploring its subtree, maintaining the correct path state as we explore other branches. This way, we efficiently explore all paths without needing to rebuild the entire path for each leaf.

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

Solution Approach

The implementation uses DFS with backtracking to explore all root-to-leaf paths and find the lexicographically smallest leaf-to-root string.

Key Components:

1. Initialization:

   • Set ans = chr(ord('z') + 1) as a sentinel value that's guaranteed to be larger than any valid string we'll encounter
   • This ensures the first valid leaf-to-root string will replace it

2. DFS Function Structure:

   def dfs(root, path):
       nonlocal ans
       if root:
           # Process current node
           # Recurse on children
           # Backtrack

3. Path Building:

   • When visiting a node, convert its value to a character: chr(ord('a') + root.val)
   • Append this character to the current path list
   • The path grows as we traverse from root toward leaves

4. Leaf Detection and String Formation:

   • A leaf is identified when root.left is None and root.right is None
   • At a leaf, reverse the path to get leaf-to-root order: ''.join(reversed(path))
   • Compare with current minimum using min() which naturally handles lexicographical comparison

5. Recursive Exploration:

   • After processing the current node, recursively call dfs(root.left, path) and dfs(root.right, path)
   • The same path list is passed to both calls, maintaining the current path state

6. Backtracking:

   • After exploring both children, remove the current node from the path: path.pop()
   • This ensures the path correctly represents the route to other nodes when we return to the parent

Algorithm Flow:

1. Start DFS from root with an empty path
2. At each node, add its character to the path
3. If it's a leaf, create the reversed string and update the minimum if needed
4. Explore left subtree (if exists)
5. Explore right subtree (if exists)
6. Remove current character from path (backtrack)
7. Return the smallest string found

The beauty of this approach is that the path list automatically maintains the correct sequence of nodes from root to the current position, and backtracking ensures we can reuse this same list for all paths without creating new lists for each traversal.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Consider this binary tree:

       0 (a)
      /     \
    1 (b)   2 (c)
   /
 3 (d)

We want to find the lexicographically smallest string from any leaf to the root.

Step-by-step DFS traversal:

1. Start at root (0/'a')

   • path = ['a']
   • Not a leaf, continue exploring

2. Go left to node (1/'b')

   • path = ['a', 'b']
   • Not a leaf (has left child), continue exploring

3. Go left to node (3/'d')

   • path = ['a', 'b', 'd']
   • This IS a leaf!
   • Reverse path to get leaf-to-root: "dba"
   • Update ans = "dba" (was initially chr(ord('z') + 1))
   • No children to explore

4. Backtrack to node (1/'b')

   • Pop 'd' from path: path = ['a', 'b']
   • No right child to explore

5. Backtrack to root (0/'a')

   • Pop 'b' from path: path = ['a']
   • Now explore right subtree

6. Go right to node (2/'c')

   • path = ['a', 'c']
   • This IS a leaf!
   • Reverse path to get leaf-to-root: "ca"
   • Compare with current ans = "dba"
   • Since "ca" < "dba" lexicographically, update ans = "ca"
   • No children to explore

7. Backtrack to root (0/'a')

   • Pop 'c' from path: path = ['a']
   • Done exploring all paths

Final answer: "ca"

The two possible leaf-to-root strings were:

• From leaf 3 to root: "dba"
• From leaf 2 to root: "ca"

Since "ca" comes before "dba" in dictionary order (comparing first character: 'c' < 'd'), the answer is "ca".

The key insight is how the path list maintains the current route and how backtracking (popping from the path) allows us to efficiently explore all branches using the same list.

Solution Implementation

Time and Space Complexity

Time Complexity: O(N * L)

The algorithm performs a depth-first search (DFS) traversal of the binary tree, visiting each node exactly once, which takes O(N) time where N is the number of nodes in the tree. At each leaf node, we perform string operations:

• ''.join(reversed(path)) creates a string from the path, which takes O(L) time where L is the length of the path (height of the tree)
• The comparison min(ans, ''.join(reversed(path))) takes O(L) time in the worst case

Since we perform these O(L) operations at each leaf node, and there can be up to O(N/2) leaf nodes in a binary tree, the total time complexity is O(N * L). In the worst case of a skewed tree, L = N, making the time complexity O(N²). In the best case of a balanced tree, L = log(N), making it O(N * log(N)).

Space Complexity: O(L) or O(H)

The space complexity consists of:

• The recursion call stack, which goes as deep as the height of the tree: O(H) where H is the height
• The path list that stores characters along the current path: O(L) where L is the current path length
• The ans string variable: O(L) for storing the smallest string

Since the path length L equals the height H at maximum, and we reuse the same path list throughout the traversal (adding and removing elements), the overall space complexity is O(H). In the worst case of a skewed tree, this becomes O(N), and in the best case of a balanced tree, it's O(log(N)).

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

Common Pitfalls

1. Incorrect String Comparison Without Reversal

The Pitfall: A common mistake is forgetting to reverse the path when comparing strings, or reversing at the wrong time. Some developers might try to build the string in leaf-to-root order during traversal or compare the root-to-leaf string directly.

Wrong Implementation:

# INCORRECT - Comparing root-to-leaf instead of leaf-to-root
if node.left is None and node.right is None:
    current_string = ''.join(path)  # Missing reversal!
    self.smallest_string = min(self.smallest_string, current_string)

Why It's Wrong: The problem specifically asks for strings formed from leaf to root, not root to leaf. Without reversal, you'd be finding the lexicographically smallest root-to-leaf path instead.

Correct Solution: Always reverse the path when at a leaf node before comparison:

if node.left is None and node.right is None:
    leaf_to_root_string = ''.join(reversed(path))
    self.smallest_string = min(self.smallest_string, leaf_to_root_string)

2. Creating New Path Lists for Each Recursive Call

The Pitfall: Passing a new copy of the path to each recursive call instead of using backtracking with a shared list.

Wrong Implementation:

def dfs(node, path):
    if not node:
        return
  
    current_char = chr(ord('a') + node.val)
    new_path = path + [current_char]  # Creating new list
  
    if node.left is None and node.right is None:
        leaf_to_root = ''.join(reversed(new_path))
        self.smallest_string = min(self.smallest_string, leaf_to_root)
  
    dfs(node.left, new_path)   # Passing copies
    dfs(node.right, new_path)
    # No backtracking needed but inefficient

Why It's Problematic: While this works correctly, it's inefficient in terms of space complexity. Creating new lists for each recursive call uses O(N * H) extra space where N is the number of nodes and H is the height of the tree.

Correct Solution: Use a single list with backtracking:

def dfs(node, path):
    if not node:
        return
  
    current_char = chr(ord('a') + node.val)
    path.append(current_char)  # Modify shared list
  
    # ... process leaf ...
  
    dfs(node.left, path)   # Pass same list
    dfs(node.right, path)
  
    path.pop()  # Backtrack - crucial step!

3. Forgetting to Handle the Backtracking Step

The Pitfall: Forgetting to remove the current character from the path after exploring both children.

Wrong Implementation:

def dfs(node, path):
    if not node:
        return
  
    path.append(chr(ord('a') + node.val))
  
    if node.left is None and node.right is None:
        leaf_to_root = ''.join(reversed(path))
        self.smallest_string = min(self.smallest_string, leaf_to_root)
  
    dfs(node.left, path)
    dfs(node.right, path)
    # MISSING: path.pop() - forgot to backtrack!

Why It's Wrong: Without backtracking, the path will contain characters from previously visited branches when exploring new branches, leading to incorrect strings that don't represent actual root-to-leaf paths.

Example of the Bug: For a tree with root 'a', left child 'b', and right child 'c' (both leaves):

• After visiting left leaf 'b', path = ['a', 'b']
• Without pop(), when visiting right child 'c', path becomes ['a', 'b', 'c']
• The string for right leaf would incorrectly be 'cba' instead of 'ca'

Correct Solution: Always include the backtracking step:

dfs(node.left, path)
dfs(node.right, path)
path.pop()  # Essential for maintaining correct path state