It’s a matter of scope. There probably aren’t many problems with your algorithm. The main thing is that you need to know that whenpassing parameters to a function, especially when passing in variable parameters (in your case, it’s a list), you have to keep it in mind. Countless. Yourproblem here is mainly focused on tmp. The reason why the state of the left subtree is retained is because when you traverse the left subtree, you add the left subtree to tmp, and then you make the next recursive call Put the added list into the list. If there is only a left subtree, there is no problem. If there is a right subtree, there will be a problem. My language expression ability is limited, so I will post the modified code for you to see

import copy class TreeNode: def __init__(self, val): self.val = val self.left, self.right = None, None def dfs(node, result, tmp=list()): if node is None: return tmp.append(node) # 这里需要用拷贝而不是用 = 赋值，也可以遍历赋值 tmp1 = copy.deepcopy(tmp) if node.left is None and node.right is None: result.append([i.val for i in tmp]) return if node.left is not None: dfs(node.left, result, tmp) # 遍历右子树需要带上不同的变量，否则左子树的tmp和右子树的tmp都指向一块内存 if node.right is not None: dfs(node.right, result, tmp1) if __name__ == '__main__': node1 = TreeNode('a') node2 = TreeNode('b') node3 = TreeNode('c') node4 = TreeNode('d') node5 = TreeNode('e') node6 = TreeNode('f') node7 = TreeNode('g') node1.left = node2 node1.right = node3 node2.left = node4 node2.right = node5 node4.left = node6 node3.left = node7 r = [] dfs(node1, result=r) print(r)