Binary Tree Paths

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"]

这道题给我们一个二叉树，让我们返回所有根到叶节点的路径，跟之前那道Path Sum很类似，比那道稍微简单一些，不需要计算路径和，只需要无脑返回所有的路径即可，那么思路还是用DFS来解

递归（dfs）

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

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        result = []
        if not root:
            return result

        def dfs(root, path):
            if not root.left and not root.right:
                result.append(path)
            if root.left:
                dfs(root.left, path+'->'+str(root.left.val))
            if root.right:
                dfs(root.right, path+'->'+str(root.right.val))
        dfs(root, str(root.val))
        return result

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

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        result = []
        if not root:
            return result
        return self.dfs(root, str(root.val), result)

    def dfs(self, root, path, result):
        if not root.left and not root.right:
            result.append(path)
        if root.left:
            self.dfs(root.left, path+'->'+str(root.left.val), result)
        if root.right:
            self.dfs(root.right, path+'->'+str(root.right.val), result)
        return result

DFS+stack:

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

class Solution(object):
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        # dfs+stack
        if not root:
            return []
        result = []
        stack = [(root, str(root.val))]
        while stack:
            node, path = stack.pop()
            if not node.left and not node.right:
                result.append(path)
            if node.right:
                stack.append((node.right,path+'->'+str(node.right.val)))
            if node.left:
                stack.append((node.left, path+'->'+str(node.left.val)))
        return result

时间

递归:O(n)

空间 O(number of leaves)?

每个 leaf 都代表着一个 path，多少个 leaf 最后就有多少个