二叉树路径-递归/非递归

在这里插入图片描述
DFS–注意"->"放后面,不然第一个会出错~~细节

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:

        out = []
        def dfs(root, res):
            if not root:
                return 
            if not root.left and not root.right:
                out.append(res+"%s"%root.val)
                return 
            if root.right:
                dfs(root.right, res+"%s->"%root.val)
            if root.left:
                dfs(root.left, res+"%s->"%root.val)
        dfs(root, "")
        return out

非递归–BFS

class Solution:
    def binaryTreePaths(self, root: TreeNode) -> List[str]:
        paths = list()
        if not root:
            return paths

        node_queue = collections.deque([root])
        path_queue = collections.deque([str(root.val)])

        while node_queue:
            node = node_queue.popleft()
            path = path_queue.popleft()

            if not node.left and not node.right:
                paths.append(path)
            else:
                if node.left:
                    node_queue.append(node.left)
                    path_queue.append(path + '->' + str(node.left.val))
                
                if node.right:
                    node_queue.append(node.right)
                    path_queue.append(path + '->' + str(node.right.val))
        return paths

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转载自blog.csdn.net/caihuanqia/article/details/114631379