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Preorder to Postorder

Tags : tree, bst, stack, geeksforgeeks, cpp, medium

Problem Statement - link #

Given an array arr[] of N nodes representing preorder traversal of BST. The task is to print its postorder traversal. In Pre-Order traversal, the root node is visited before the left child and right child nodes. Post-order traversal is one of the multiple methods to traverse a tree.

Your Task: You need to complete the given function and return the root of the tree. The driver code will then use this root to print the post order traversal.

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(N)

Examples #

Example 1:

Input:
N = 5
arr[]  = {40,30,35,80,100}
Output: 35 30 100 80 40
Explanation: PreOrder: 40 30 35 80 100
InOrder: 30 35 40 80 100
Therefore, the BST will be:
              40
           /      \
         30       80
           \        \   
           35      100
Hence, the postOrder traversal will
be: 35 30 100 80 40

Example 2:

Input:
N = 8
arr[]  = {40,30,32,35,80,90,100,120}
Output: 35 32 30 120 100 90 80 40

Constraints #

Solutions #

class Solution
{
    public:
    //Function that constructs BST from its preorder traversal.
    Node* post_order(int pre[], int size)
    {
        //code here
        Node* root = NULL;
        if(!size) return root;
        root = newNode(pre[0]);
        int i=1;
        stack<Node*> s;
        Node* temp = root;
        
        while(i<size){
            if(pre[i] < temp->data){
                temp->left = newNode(pre[i++]);
                s.push(temp);
                temp = temp->left;
            }
            else {
                if(s.empty() || (pre[i] < s.top()->data)){
                    temp->right = newNode(pre[i++]);
                    temp = temp->right;
                }else {
                    temp = s.top(); s.pop();
                }
            }
        }
        return root;
    }
};