257. Binary Tree Paths

Problem Description

The problem presents us with a binary tree, where each node contains an integer value and can have a left and/or right child. Our goal is to find all the unique paths from the root of the binary tree to its leaves. A leaf is defined as a node that has no children, which means neither a left nor a right child. The paths should be represented as strings, with each node's value on the path concatenated by "->". For instance, if the path goes through nodes with values 1, 2, and 3 in that order, the path string should be "1->2->3". We are required to return these path strings in any order.

Intuition

To solve this problem, we can utilize a technique called Depth-First Search (DFS). This strategy explores as far as possible along each branch before backtracking. Here's how we arrive at the solution approach:

• We can start at the root and perform a DFS traversal on the binary tree.
• As we traverse, we keep track of the current path by noting the nodes visited so far in the sequence.
• Whenever we reach a leaf node (a node without children), we record the path we've taken to get there. This is a complete root-to-leaf path, so we add this to our list of answers.
• The key part of this process is backtracking. When we visit a node and explore all of its children, we backtrack, which means we remove the node from our current path and return to the previous node to explore other paths.
• We continue this process until all paths are explored and we have visited all the leaves.

The recursive function dfs() in the solution code is where the DFS takes place. It takes the current node as a parameter, and as the function is called recursively, a local variable t keeps track of the current path as a list of node values. If a leaf node is reached, the current path is converted to the path string format required and added to the list ans, which contains all the full path strings. After exploring a node's left and right children, the node's value is popped from the path list to allow backtracking to the previous node's state.

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

Solution Approach

The solution approach for the given problem utilizes a typical pattern for tree traversal problems, which is the Depth-First Search (DFS) algorithm. Here's a step-by-step explanation of how the solution is implemented:

1. Define a Recursive Function (dfs): The recursive function dfs is defined within the class Solution. It is invoked with the current node being visited. This function does not return any value but instead updates the ans list with the path strings.

2. Base Case for Recursion: In the function dfs, before going further into recursion, we check if the given root node is None, indicating that we've reached beyond a leaf node. In this case, the function simply returns without performing any further action.

3. Track the Current Path: A local variable t in the class scope is used to keep track of the current path. It's a stack (implemented using a list in Python) that we update as we go down the tree. For each node, we convert its value to a string and append it to t.

4. Check for Leaf Node: In the DFS process, if the current node is a leaf (both left and right child nodes are None), we convert the current path into the required string format, which is obtained by joining the sequence of node values in t with "->". This complete path string is added to the ans list.

5. Recursive DFS Calls: If the current node has children, we perform DFS for both the left child (if not None) and right child (if not None). The function calls itself with root.left and root.right correspondingly.

6. Backtracking: After exploring both children from the current node, we need to backtrack. This ensures that when we return from the recursive call, the current path t reflects the state as if the current node was never visited. We achieve this by popping the last value from the t.

7. Return Paths: Once the DFS is complete, the ans list will be populated with all the path strings from root to all the leaf nodes. This list is returned as the final result of the binaryTreePaths method.

The use of a stack for tracking the paths and the pattern of adding to the list only at leaf nodes are critical parts of the algorithm. The recursive DFS makes sure all potential paths are explored, while backtracking ensures that every unique path is properly captured without duplication.

Example Walkthrough

Let's illustrate the solution approach using a small example binary tree:

Consider the following binary tree:

1     1
2    / \
3   2   3
4  /     \
5 5       4

Here's how we would apply the solution approach:

1. We start with the root node which is 1.

2. We initialize t with the value of the current node: t = ["1"].

3. The node 1 is not a leaf, so we continue. We have two recursive DFS calls to make because both left and right children exist: root.left (which is 2) and root.right (which is 3).

4. First DFS call: We move to the left child which is 2 and update t to ["1", "2"].

5. Node 2 has a left child but no right child. We make a recursive DFS call to the left child which is 5.

6. On reaching the node 5, we update t to ["1", "2", "5"].

7. Node 5 is a leaf node, so we join the elements of t with "->" to form the path string "1->2->5". We add this to our ans list.

8. Backtracking: We finished visiting 5, so we pop it from t making t = ["1", "2"].

9. There are no more children for 2 to visit, so we pop it from t as well. Now t = ["1"].

10. Second DFS call: Now, we explore the right child of 1, which is 3. We update t to ["1", "3"].

11. Node 3 has a right child 4 but no left child. We make a recursive DFS call to the right child which is 4.

12. On reaching 4, we update t to ["1", "3", "4"].

13. Node 4 is a leaf node, so we join the elements of t with "->" to form the path string "1->3->4". We add this to our ans list.

14. Backtracking: We've visited 4, so we pop it from t, reverting t back to ["1", "3"].

15. After exploring node 3's right child, we backtrack one more time, popping 3 from t and t is now back to ["1"].

Finally, since all paths have been explored, our ans list contains the root-to-leaf path strings: ["1->2->5", "1->3->4"]. We'd return this list as the output of our function.

Solution Implementation

Time and Space Complexity

The given Python function binaryTreePaths traverses a binary tree to find all root-to-leaf paths. The primary operation is a Depth-First Search (DFS) defined in a nested function. Here's the analysis of its complexities:

Time Complexity

The time complexity is O(N), where N is the number of nodes in the tree. This is because each node in the binary tree is visited exactly once during the depth-first search. Therefore, the function's time complexity is directly proportional to the number of nodes.

Space Complexity

The space complexity of the function is also O(N). The major factors contributing to the space complexity are:

1. The recursive call stack of the DFS, which in the worst case could be O(N) when the binary tree degrades to a linked list.
2. The list t that holds the current path. In the worst case, when the binary tree is completely skewed (e.g., each parent has only one child), this list can also take up to O(N) space.
3. The list ans that contains all the paths. In the best case, where each node has two children, there will be N/2 leaf nodes resulting in N/2 paths. Although each path could be of varying length, the concatenation of all paths will still take O(N) space since each node's value is only included in one path string.

Considering the stack space for recursive calls and the space for storing the paths, the space complexity in the worst case scenario will be linear with respect to the number of nodes in the tree.

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