Code360 powered by Coding Ninjas X Naukri.com. Code360 powered by Coding Ninjas X Naukri.com
Last Updated: 3 Aug, 2017

Preorder Binary Tree

Easy
Asked in company
Walmart

Problem statement

You are given a ‘Binary Tree’.


Return the preorder traversal of the Binary Tree.


Example:
Input: Consider the following Binary Tree:

Example

Output: 
Following is the preorder traversal of the given Binary Tree: [1, 2, 5, 3, 6, 4]


Input Format:
The only line contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image will be:

alt text

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null(-1).

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1


Output Format:
Return an array representing the preorder traversal of the given binary tree.


Note :
You do not need to print anything; it has already been taken care of. Just implement the given function.

Approaches

01 Approach

Approach:

In preorder traversal of a binary tree, we visit the current node, then the left subtree and finally the right subtree. So we can recursively solve this problem by storing the value of current node then visiting the left subtree then the right subtree. 


 

The steps are as follows:

function preOrderHelper(TreeNode<int>* cur, vector<int> path)

  1. if(cur == null)
    1. return
  2. push value of current node to the path
  3. preOrderHelper(cur->left,path)
  4. preOrderHelper(cur->right,path)


 

function preOrder(TreeNode<int>* root)

  1. Initialise empty array path
  2. preOrderHelper(root,path)
  3. return path