Last Updated: 3 Aug, 2017

Preorder Binary Tree

Easy

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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.

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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)

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

function preOrder(TreeNode<int>* root)

Initialise empty array path
preOrderHelper(root,path)
return path

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02 Approach

Approach:

We can append the value of the current node to the ‘path’ array and then visit the left subtree. Problem is, while iterating through the tree we won’t have a way to go back to the right subtree after traversing the left subtree. To solve this, we can make use of a stack. When visiting a node, we will do the following:

Pop the current node from the stack.
If a right subtree exists, push it into the stack to return to later.
If a left subtree exists, push it into the stack.

Since the left subtree is at the top of the stack we will visit it first. After we have visited all the nodes in the left subtree, the right subtree will be at the top,and we will visit it.

The steps are as follows:

function preOrder(TreeNode<int>* root)

Initialise empty array path
Initialise empty stack ‘s’
s.push(root)
while(!s.empty())
1. cur = s.pop()
2. Push value of current node to the path
3. if(cur->right)
  1. s.push(cur->right)
4. if(cur->left)
  1. s.push(cur->left)
return path

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