Suppose we have a binary tree. our task is to create an inverted binary tree. So if the tree is like below −
The inverted tree will be like
To solve this, we will use a recursive approach
- if the root is null, then return
- swap the left and right pointers
- recursively solve left subtree and right subtree
Example (Python)
Let us see the following implementation to get a better understanding −
class TreeNode: def __init__(self, data, left = None, right = None): self.data = data self.left = left self.right = right def make_tree(elements): Tree = TreeNode(elements[0]) for element in elements[1:]: insert(Tree, element) return Tree def height(root): if root is None: return 0 else : # Compute the height of left and right subtree l_height = height(root.left) r_height = height(root.right) #Find the greater one, and return it if l_height > r_height : return l_height+1 else: return r_height+1 def print_given_level(root, level): if root is None: return if level == 1: print(root.data,end = ',') elif level > 1 : print_given_level(root.left , level-1) print_given_level(root.right , level-1) def level_order(root): h = height(root) for i in range(1, h+1): print_given_level(root, i) def insert(temp,data): que = [] que.append(temp) while (len(que)): temp = que[0] que.pop(0) if (not temp.left): if data is not None: temp.left = TreeNode(data) else: temp.left = TreeNode(0) break else: que.append(temp.left) if (not temp.right): if data is not None: temp.right = TreeNode(data) else: temp.right = TreeNode(0) break else: que.append(temp.right) class Solution(object): def invertTree(self, root): """ :type root: TreeNode :rtype: TreeNode """ self.solve(root) return root def solve(self,root): if not root: return temp = root.left root.left = root.right root.right = temp self.solve(root.left) self.solve(root.right) tree1 = make_tree([1,2,2,3,4,None,3]) ob1 = Solution() tree2 = ob1.invertTree(tree1) level_order(tree2)
Input
[1,2,2,3,4,None,3]
Output
1,2,2,3,None,4,3,