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Binary Tree Paths

TreesDFS

Problem Statement

Given the `root` of a binary tree, return all root-to-leaf paths in any order. A leaf is a node with no children.

Example

Output: ["1→2→5", "1→3"]

Algorithm Explanation

We can solve this problem using a depth-first search (DFS) approach. We'll traverse the tree from the root, keeping track of the current path.

Algorithm Steps

DFS Function: Create a recursive DFS function that takes a node and the current path string as arguments.
Path String: At each node, append its value to the current path string.
Leaf Node: If the current node is a leaf, we've found a complete path. Add the path string to our result list.
Recursive Calls: If the node is not a leaf, make recursive calls for its left and right children, passing along the updated path string.

Start

Starting DFS from the root.

Current Path

Result

Binary Tree Paths Solution


class Solution:
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        result = []
        
        def dfs(node, path):
            if not node:
                return
            
            path += str(node.val)
            if not node.left and not node.right:
                result.append(path)
            else:
                path += "->"
                dfs(node.left, path)
                dfs(node.right, path)

        dfs(root, "")
        return result