Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes’ values.

Given binary tree

   1
    \
     2
    /
   3

return [3,2,1].

Using recursion is not allowed.

Method:

pre-order traversal is root-left-right, and post order is left-right-root. modify the code for pre-order to make it root-right-left, and then reverse the output so that we can get left-right-root.

Solution:

Time: O(n)
Space: O(n)

/**
 * Definition for binary tree
 * class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) {
 *      val = x;
 *      left=null;
 *      right=null;
 *     }
 * }
 */
public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode A) {
        ArrayList<Integer> result = new ArrayList<>();
        if (A == null) return result;
        Deque<TreeNode> stack = new ArrayDeque<>();
        stack.push(A);
        while (!stack.isEmpty()) {
            TreeNode curr = stack.pop();
            result.add(curr.val);
            if (curr.left != null) {
                stack.push(curr.left);
            }
            if (curr.right != null) {
                stack.push(curr.right);
            }
        }
        Collections.reverse(result);
        return result;
    }
}