Program to perform an Inorder Traversal of a binary tree in Python

Suppose we have a binary tree; we have to find a list that contains the inorder traversal of root as a list. As we know the inorder traversal is a way of traversing all nodes in a tree where we −

• Recursively traverse the left subtree.

• Traverse the current node.

• Recursively traverse the right subtree.

We have to try to solve this problem in iterative fashion.

So, if the input is like

then the output will be [12,13,4,16,7,14,22]

To solve this, we will follow these steps −

• inorder := a new list

• stack := an empty stack

• Do the following infinitely, do

  • if root is not null, then

    • push root into the stack

    • root := left of root

  • otherwise when stack is not empty, then

    • root := top element of stack and pop from stack

    • insert value of root at the end of inorder

    • root := right of root

  • otherwise,

    • come out from the loop

• return inorder

Let us see the following implementation to get better understanding −

Example

class TreeNode:
   def __init__(self, value):
      self.val = value
      self.left = None
      self.right = None
class Solution:
   def solve(self, root):
      inorder = []
      stack = []
      while True:
         if root:
            stack.append(root)
            root = root.left
         elif stack:
            root = stack.pop()
            inorder.append(root.val)
            root = root.right
         else:
            break
      return inorder

ob = Solution()
root = TreeNode(13)
root.left = TreeNode(12)
root.right = TreeNode(14)
root.right.left = TreeNode(16)
root.right.right = TreeNode(22)
root.right.left.left = TreeNode(4)
root.right.left.right = TreeNode(7)
print(ob.solve(root))

Input

root = TreeNode(13)
root.left = TreeNode(12)
root.right = TreeNode(14)
root.right.left = TreeNode(16)
root.right.right = TreeNode(22)
root.right.left.left = TreeNode(4)
root.right.left.right = TreeNode(7)

Output

[12, 13, 4, 16, 7, 14, 22]