Suppose we have a binary tree. We have to traverse this tree using the inorder traversal scheme without using recursion. So if the tree is like
Then the traversal will be [2,5,7,10,15,20]
To solve this, we will follow these steps −
- Create two array res and stack, set curr := root
- Run one infinite loop
- while current is not null
- push curr into a stack, and set curr := left of curr
- when the length of stack = 0, then return res
- node := popped element from the stack
- insert a value of node into res
- curr := right of curr
- while current is not null
Example
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 insert(temp,data): que = [] que.append(temp) while (len(que)): temp = que[0] que.pop(0) if (not temp.left): temp.left = TreeNode(data) break else: que.append(temp.left) if (not temp.right): temp.right = TreeNode(data) break else: que.append(temp.right) def make_tree(elements): Tree = TreeNode(elements[0]) for element in elements[1:]: insert(Tree, element) return Tree class Solution(object): def inorderTraversal(self, root): res, stack = [], [] current = root while True: while current: stack.append(current) current = current.left if len(stack) == 0: return res node = stack[-1] stack.pop(len(stack)-1) if node.data != None: res.append(node.data) current = node.right return res ob1 = Solution() root = make_tree([10,5,15,2,7,None,20]) print(ob1.inorderTraversal(root))
Input
[10,5,15,2,7,null,20]
Output
[2,5,7,10,15,20]