BinaryTreePostorderTraversal.java

package tree;
import java.util.*;
/**
 * Created by gouthamvidyapradhan on 28/07/2018.
 * Given a binary tree, return the postorder traversal of its nodes' values.

 Example:

 Input: [1,null,2,3]
 1
 \
 2
 /
 3

 Output: [3,2,1]
 Follow up: Recursive solution is trivial, could you do it iteratively?

Solution: O(N). Maintain a stack, for every node which you pop from stack add it to result list, push left
 and right node to stack. Reverse the result list and return this as the answer.
 */
public class BinaryTreePostorderTraversal {
    public static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }

    public static void main(String[] args) throws Exception{
        TreeNode root = new TreeNode(1);
        root.right = new TreeNode(2);
        root.right.left = new TreeNode(3);
        List<Integer> result = new BinaryTreePostorderTraversal().postorderTraversal(root);
        result.forEach(System.out::println);
    }

    public List<Integer> postorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        List<Integer> result = new ArrayList<>();
        if(root != null){
            stack.push(root);
        }
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            result.add(node.val);
            if(node.left != null){
                stack.push(node.left);
            } if(node.right != null){
                stack.push(node.right);
            }
        }
        Collections.reverse(result);
        return result;
    }



}