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)])