Inverting binary tree (recursive)

Question

I can't figure out how to output a reversed binary tree. This is what I've come up so far + my pseudo code.

Creating a binary tree

#Creating the binary tree
from binarytree import build
from binarytree import tree
   
# List of nodes 
nodes =[4, 2, 7, 1, 3, 6, 9] 
  
# Builidng the binary tree 
binary_tree = build(nodes) 
print('Binary tree from example :\n ', 
      binary_tree) 
  
# Getting list of nodes from 
# binarytree 
print('\nList from binary tree :',  
      binary_tree.values) 

Output:

Pseudo code:

#def invert_tree(nodes, root)

#Stop recursion if tree is empty

#swap left subtree with right subtree

#invert left subtree

#invert right subtree

Answer 1

I found an answer

nodes = [[4], [7, 2], [1, 3, 6, 9]]

Recursive

newlist1 = []
def reverse_tree(lst):
    if not lst:
        return
    newlist1.append(lst.pop(0)[::-1])
    reverse_tree(lst)

reverse_tree(nodes)

print(newlist1)

Output:

[[4], [2, 7], [9, 6, 3, 1]]

Using list comprehension

#ListComprehension
newlist2 = [x[::-1] for x in nodes]
print(newlist2)

Output:

[[4], [2, 7], [9, 6, 3, 1]]

Answer 2

Your question isn't clear;

From what I understood:

To print the nodes of an inverted tree:
    Try Level order traversal, more precisely you can use the BFS method.

OR: If you want to invert a Binary tree; My approach (given the root node of the tree):

def invert_tree(self, root):
    if root:
        temp=root.left
        root.left = self.invert_tree(root.right)
        root.right = self.inver_tree(temp)
        return root

Since, every node in the tree in visited once, the time complexity is O(n).

Answer 3

TreeNode* Invert(TreeNode* tree) {

if(tree == NULL) return NULL;

TreeNode* temp;
temp = tree->left;
tree->left = tree->right;
tree->right = temp;

Invert(tree->left);
Invert(tree->right);
return tree;

}