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

Note: A leaf is a node with no children.

Example:

Input:   1 /   \2     3 \  5Output: ["1->2->5", "1->3"]Explanation: All root-to-leaf paths are: 1->2->5, 1->3

思路:看了discuss里面的提示,写了这个方法.思路比较清晰,用递归的方法,先列出终止条件1.root为空 2.没有左右孩子(叶子节点)

有问题的是在于两个递归一个用'=',一个用'+='.这是因为需要返回所有root到leaf的路径,如果两个都是'='的话,只能返回一条路径.

# Definition for a binary tree node.# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    def binaryTreePaths(self, root):        """        :type root: TreeNode        :rtype: List[str]        """        if not root:            return []        if not root.left and not root.right:            return [str(root.val)]        path = [str(root.val)+'->' + p for p in self.binaryTreePaths(root.left)]        path += [str(root.val)+'->' + p for p in self.binaryTreePaths(root.right)]        return path

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