Leetcode 257. Binary Tree Paths

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"]

本题可以用DFS或者BFS。

解法一： DFS

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
            
        result = []
        self.dfs(root, result, [str(root.val)])
        return result
        
    def dfs(self, root, result, path):
        if root.left is None and root.right is None:
            result.append('->'.join(path))
            return
        
                
        if root.right:
            path.append(str(root.right.val))
            self.dfs(root.right, result, path)
            path.pop()
        if root.left:
            path.append(str(root.left.val))
            self.dfs(root.left, result, path)
            path.pop()

 

这个解法中需要注意的是current path的初始值是[str(root.val)]而不是。这个很类似的解法使用了pop。

 

如果要不使用pop的话

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
            
        result = []
        self.dfs(root, result, [str(root.val)])
        return result
        
    def dfs(self, root, result, path):
        if root.left is None and root.right is None:
            result.append('->'.join(path))
            return
        
                
        if root.right:
            self.dfs(root.right, result, path+[str(root.right.val)])
        if root.left:
            self.dfs(root.left, result, path+[str(root.left.val)])