深度优先 广度优先

深度优先：首先遍历最低层的节点，然后逐步上移到根节点。
dfs + recursion过于的慢
尽量不使用递归
例

257. Binary Tree Paths
     Easy
     Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

1
/
2 3

5

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

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

自己写的递归

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

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        if root is None:
            return []
        ls,res='',[]
        self.dfs(root,ls,res)
        return res
        
    def dfs(self,node,ls,res):
        if node.left:
            self.dfs(node.left, ls+str(node.val)+'->', res)
        if node.right:
            self.dfs(node.right, ls+str(node.val)+'->', res)
        if node.left is None and node.right is None: 
            res.append(ls+str(node.val))

别人写的用栈不用递归

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#用stack代替递归，会快一些嘞
class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        if not root:
            return []
        res,stack=[],[(root,"")]
        while stack:
            node, ls = stack.pop()
            if not node.left and not node.right:
                res.append(ls+str(node.val))
            if node.right:
                stack.append((node.right, ls+str(node.val)+"->"))
            if node.left:
                stack.append((node.left, ls+str(node.val)+"->"))
        return res

二叉树的最大路径和 #124
找不到快一点的方法 有时间再做一次