Suppose we have two binary trees; we have to check whether the sequence of leaves left-to-right in both trees are the same.
So, if the input is like
then the output will be True as the sequence is [2, 6] for both trees.
To solve this, we will follow these steps:
- c := a new list
- Define a function inorder() . This will take root, and c
- if c is null, then
- c := a new list
- if root is not null, then
- inorder(left of root, c)
- if left of root is null and right of root is null, then
- insert value of root at the end of c
- inorder(right of root, c)
- return c
- From the main method, do the following:
- if inorder(root0) is same as inorder(root1), then
- return True
- otherwise,
- return False
Let us see the following implementation to get better understanding:
Example
class TreeNode: def __init__(self, data, left = None, right = None): self.val = data self.left = left self.right = right class Solution: c = [] def inorder(self, root, c=None): if c is None: c = [] if root: self.inorder(root.left, c) if not root.left and not root.right: c.append(root.val) self.inorder(root.right, c) return c def solve(self, root0, root1): if self.inorder(root0) == self.inorder(root1): return True else: return False ob = Solution() root1 = TreeNode(1) root1.right = TreeNode(3) root1.right.left = TreeNode(2) root1.right.right = TreeNode(6) root2 = TreeNode(1) root2.left = TreeNode(3) root2.right = TreeNode(6) root2.left.left = TreeNode(2) print(ob.solve(root1, root2))
Input
root1 = TreeNode(1) root1.right = TreeNode(3)root1.right.left = TreeNode(2)
root1.right.right = TreeNode(6)
root2 = TreeNode(1)
root2.left = TreeNode(3)
root2.right = TreeNode(6)root2.left.left = TreeNode(2)
Output
True