# Binary Tree Paths

## 1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?
• Can the input tree be Null?
• Yes, it can be. In that case there are no root-to-leaf paths, correct?
• Can a root-to-leaf path be the root itself, if it has no children?
• Yes, that can occur.
``````HAPPY CASE
Input:     1
/   \
2     3
\
5

Output: ["1->2->5", "1->3"]

Input:     1
/   \
2     3
\     \
5     2

Output: ["1->2->5", "1->3->2"]

EDGE CASE (Only Root Node)
Input:  1
Output: ["1"]
``````

## 2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

If you are dealing with Binary Trees some common techniques you can employ to help you solve the problem:

• Think about appropriate Tree Traversal: Pre-Order, In-Order, Post-Order, Level-Order
• Choosing a specific tree traversal that follows a general root-to-leaf path should help us identify all of the possible routes.
• Store nodes within a HashMap to refer to later
• We don’t have a specific way of referring to previous nodes in a path that could be used in a HashMap. So, a HashMap would not help us as much in this context.
• Using Binary Search to find an element
• We are not working with a Binary Search Tree.
• Applying a level-order traversal with a queue
• Using this approach may complicate our code

## 3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

SOLUTION 1:

General Idea: Pre-Order traversal through the Binary Tree while keeping track of a current path. When we reach a leaf node, store the current path and pop elements off as we backtrack and explore more of the tree.

``````0) Create helper function to allow us to retain memory of allPaths
1) Basecase: If the current element is Null, return
2) Recursive: Build Current Path
a) Add the current node to the current path data structure
b) If the current node is a leaf node, store the current path in the tree
c) Recurse Left and Recurse Right
d) Pop the current node off the current path, since we will no longer have more root-to-leaf paths that go through the current node.
3) Create allPaths array to retain memory of allPaths
4) Run helper function to collect allPaths
5) Return allPaths``````

SOLUTION 2:

General Idea: Lets build the paths as we progress through the nodes. If the node is a leaf node, then add it to our results.

``````1. Create a helper function to recursively progress through the nodes.
a. Collect the values and build string
b. At a leaf node add built string to results
c. Progress to left node and right node
2. Create results array
3. Call helper function to build results
4. Return results ``````

⚠️ Common Mistakes

• Not noticing the need for a helper function to retain memory
• Choosing the wrong traversal type
• Try to walk through the problem by hand and see the order in which you are processing the nodes. This will clue you into the type of traversal necessary
• Suppose you choose post order traversal, we would go left - right - check. No root node path.
• Suppose you choose in order traversal, we would go left - check - right. No root node for path.

## 4: I-mplement

Implement the code to solve the algorithm.

Solution 1

``````class Solution:
def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:

# Create helper function to allow us to retain memory of allPaths
def helper(root, currPath):
# Basecase: If the current element is Null, return
if not root:
return

# Recusive: Build Current Path
# Add the current node to the current path data structure
currPath.append(str(root.val))

# If the current node is a leaf node, store the current path in the tree
if not root.left and not root.right:
allPaths.append("->".join(currPath))

# Recurse Left and Recurse Right
helper(root.left, currPath)
helper(root.right, currPath)

# Pop the current node off the current path, since we will no longer have more root-to-leaf paths that go through the current node.
currPath.pop()

# Create allPaths array to retain memory of allPaths
allPaths = []

# Run helper function to collect allPaths
helper(root, [])

# Return allPaths
return allPaths``````

Solution 2

``````# 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 binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
# Create a helper function to recursively progress through the nodes.
def dfs(node, s):
# Collect the values and build string
if s != "":
s += "->"
s += str(node.val)

# At a leaf node add built string to results
if not node.left and not node.right:
res.append(s)

# Progress to left node and right node
if node.left:
dfs(node.left, s)
if node.right:
dfs(node.right, s)

# Create results array
res = []

# Call helper function to build results
dfs(root, "")

# Return results
return res``````
``````class Solution {
public List<String> binaryTreePaths(TreeNode root) {
// Create results array
// Call helper function to build results
if (root != null) searchBT(root, "", answer);
// Return results
}

// Create a helper function to recursively progress through the nodes
private void searchBT(TreeNode root, String path, List<String> answer) {
// Collect the values and build string
// At a leaf node add built string to results

// Progress to left node and right node
if (root.left != null) searchBT(root.left, path + root.val + "->", answer);
if (root.right != null) searchBT(root.right, path + root.val + "->", answer);
}
}``````

## 5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors

## 6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume `N` represents the number of nodes in tree

• Solutions 1 & 2

• Time Complexity: `O(N)` because we need to visit each node in binary tree.
• Space Complexity: `O(N)` because we need to create a results array of all root-to-left paths which is `(n+1)/2` leafs. 