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

Example 1:

```Input: root = [1,null,2,3]
Output: [3,2,1]
```

Example 2:

```Input: root = []
Output: []
```

Example 3:

```Input: root = [1]
Output: [1]
```

Example 4:

```Input: root = [1,2]
Output: [2,1]
```

Example 5:

```Input: root = [1,null,2]
Output: [2,1]
```

Constraints:

• The number of the nodes in the tree is in the range `[0, 100]`.
• `-100 <= Node.val <= 100`

Recursive solution is trivial, could you do it iteratively?

## Recursive DFS Solution

```/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<>();
dfs(root, ret);
return ret;
}

private void dfs(TreeNode root, List<Integer> ret) {
if (root != null) {
dfs(root.left, ret);
dfs(root.right, ret);
}
}
}```

## Iterative DFS Solution

```/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
Stack<TreeNode> stack = new Stack<>();
if (root == null) return ans;

stack.push(root);
while (!stack.isEmpty()) {
TreeNode cur = stack.pop();
if (cur.left != null) {
stack.push(cur.left);
}
if (cur.right != null) {
stack.push(cur.right);
}
}
return ans;
}
}```
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