257. 二叉树的所有路径

代码一

class Solution(object):
    # 思路：用前序式DFS，从root到叶子。
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        if not root:
            return []
        res = []
        return self.pre_DFS(root, res, "")

    def pre_DFS(self, root, res, road):
        if root:
            road += str(root.val)
            if not root.left and not root.right:
                res.append(road)
            else:
                road += '->'
                self.pre_DFS(root.left, res, road)
                self.pre_DFS(root.right, res, road)
        return res

代码二

class Solution(object):
    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)]
        pathList = []
        if root.left:
            pathList += self.binaryTreePaths(root.left)
        if root.right:
            pathList += self.binaryTreePaths(root.right)
        for index, path in enumerate(pathList):
            pathList[index] = str(root.val) + "->" + path
        return pathList

用list返回二叉树的所有路径。

class Solution(object):
    # 前序式DFS——用list返回二叉树的所有路径
    # 注意每趟递归前要将当前节点pop()掉
    def pre_DFS(self, node, res, road):
        # 当前节点为空，直接return
        if not node:
            return
        # 否则将当前节点加入路径中
        road.append(node.val)
        # 当前节点是叶子则将路径加入外层list中
        if not node.left and not node.right:
            res.append(list(road))
        # 前序式递归当前节点的左右子树
        self.pre_DFS(node.left, res, road)
        self.pre_DFS(node.right, res, road)
        # 每趟递归前将当前节点pop()
        road.pop()
        # 返回外层list
        return res