leetcode 刷题总结

leetcode 刷题总结

题目描述
给定一个二叉树，返回所有从根节点到叶子节点的路径。
说明: 叶子节点是指没有子节点的节点。

示例
输入:
1
/ \
2 3
\
5
输出: [“1->2->5”, “1->3”]
解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

递归方法

class Solution:
    def binaryTreePaths(self, root):
        """
        :type root: TreeNode
        :rtype: List[str]
        """
        def gen(root, path):
            if root:   # 当前节点存在
                path += str(root.val)
                if not root.left and not root.right: # 当前节点是叶子节点
                    paths.append(path)
                else:
                    path += '->'  #当前节点不是叶子节点，继续递归遍历
                    gen(root.left, path)
                    gen(root.right, path)
       	paths = []
        gen(root, '')
        return paths

非递归方法

# 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]:
        paths = []
        if not root:
            return paths
       	nodedeq = collections.deque([root])
        pathdeq = collections.deque([str(root.val)])
        while nodedeq:
            node = nodedeq.popleft()
            path = pathdeq.popleft()
            
            if not node.left and not node.right:
                paths.append(path)
            else:
                if node.left:
                    nodedeq.append(node.left)
                    pathdeq.append(path + '->' + str(node.left.val))
                if node.right:
                    nodedeq.append(node.right)
                    pathdeq.append(path + '->' + str(node.right.val))
     	return paths