# 145. Binary Tree Postorder Traversal

[Link](https://leetcode.com/problems/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**

由於root node會是最後travel，那麼透過add(0, node.val)的方式，保證先travel的會在list的最後面\
接下來就是保證想先travel的節點先丟到stack，\
1\. node.left先放入stack\
2\. 接下來再放node.right

```
 public List<Integer> postorderTraversal(TreeNode root) {     
        List<Integer> ret = new LinkedList<>();
        if(root == null) return ret;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            ret.add(0, node.val);
            if(node.left != null){
                stack.push(node.left);
            }
            if(node.right != null){
                stack.push(node.right);    
            } 
        
        }
        return ret;
    }
```
