Binary Tree Paths(dfs,分治）

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

分析

注意此处返回有root == null和root是叶节点2个情况。

if not root.left and not root.right 错的一塌糊涂！！！！！

dfs带path 和ret，返回void.

分治 return root.val+L+R

dfs：返回时候考虑当前node做头和做尾的情况，子调用时候不考虑自己。

# 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):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        ret = []
        if not root:
            return ret
        self.dfs(root,ret,"")
        return list(ret)

    def dfs(self,root,ret,path):
        if not root.left and not root.right:
            path+=str(root.val)
            ret.append(path)
            return
        path+=str(root.val)+"->"
        if root.left:
            self.dfs(root.left,ret,path)
        if root.right:
            self.dfs(root.right,ret,path)
        # path.pop()

分治：结束返回只考虑自己，往下走分支时候考虑把自己加入子调用。

# 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):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        ret = []
        if not root:
            return []
        if not root.left and not root.right:
            return [str(root.val)]
        ret+=[str(root.val)+"->"+i for i in self.binaryTreePaths(root.left)]
        ret+=[str(root.val)+"->"+i for i in self.binaryTreePaths(root.right)]
        return ret