Binary Tree Postorder Traversal

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

Recursive
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
recursivePreorderTraversal(root, result);
return result;
}
void recursivePreorderTraversal(TreeNode root, List<Integer> result) {
if (root != null) {
recursivePreorderTraversal(root.left, result);
recursivePreorderTraversal(root.right, result);
result.add(root.val);
}
}
}
LeetCode Official Solution
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
LinkedList<TreeNode> stack = new LinkedList<>();
LinkedList<Integer> output = new LinkedList<>();
if (root == null) {
return output;
}
stack.add(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pollLast();
output.addFirst(node.val);
if (node.left != null) {
stack.add(node.left);
}
if (node.right != null) {
stack.add(node.right);
}
}
return output;
}
}
Iterative - Using Stack
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
LinkedList<Integer> result = new LinkedList<>();
Deque<TreeNode> stack = new ArrayDeque<>();
TreeNode p = root;
while(!stack.isEmpty() || p != null) {
if(p != null) {
stack.push(p);
result.addFirst(p.val); // Reverse the process of preorder
p = p.right; // Reverse the process of preorder
} else {
TreeNode node = stack.pop();
p = node.left; // Reverse the process of preorder
}
}
return result;
}
}