257.Binary Tree Paths

Tags: [tree], [binary_tree], [path], [recursion], [backtracking]

Com: {fb}

Link: https://leetcode.com/problems/binary-tree-paths/\#/description

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

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

Solution: Backtracking

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []

        result = []
        buff = []
        self.binary_tree_paths_helper(root, buff, result)
        return result

    def binary_tree_paths_helper(self, root, buff, result):
        # base case
        if not root:
            return
        if not root.left and not root.right:
            buff.append(root.val)
            result.append('->'.join([str(x) for x in buff]))
            buff.pop()
            return

        buff.append(root.val)
        self.binary_tree_paths_helper(root.left, buff, result)
        self.binary_tree_paths_helper(root.right, buff, result)
        buff.pop()

Revelation:

• The real base case is 'if not root.left and not root.right' at where it is a leaf node.
• We do backtracking, we first try to buff.append(), after the recursion call, we must do buff.pop().

Note:

• Time complexity = O(n), n is the number of nodes in the given tree. Because each node we only visited once.