Check if each internal node of a BST has exactly one child in Python

PythonServer Side ProgrammingProgramming

Suppose we have the preorder traversal of a binary search tree (BST). We have to check whether each internal node has only one child or not.

So, if the input is like preorder = [22, 12, 13, 15, 14], then the output will be True as BST is like −

To solve this, we can follow one efficient approach. As all decedents of a node is either smaller or larger, then we can we can follow these steps −

  • Get the next preorder successor of the node

  • Get the last preorder successor of the node

  • Now when both the successors are less than or greater than the current node then check again otherwise return false.

To solve this, we will follow these steps −

  • next := 0, last := 0
  • for i in range 0 to size of preorder - 1, do
    • next := preorder[i] - preorder[i+1]
    • last := preorder[i] - last value of preorder
    • if next * last < 0, then
      • return False
  • return True

Let us see the following implementation to get better understanding −

Example

 Live Demo

def solve(preorder):
next = 0
last = 0
for i in range(len(preorder)-1):
next = preorder[i] - preorder[i+1]
last = preorder[i] - preorder[-1]
if next * last < 0:
return False
return True
preorder = [22, 12, 13, 15, 14] print(solve(preorder))

Input

[22, 12, 13, 15, 14]

Output

True
raja
Published on 30-Dec-2020 13:52:26
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