# Posts tagged as “postorder”

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 = 
Output: 
```

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`

Idea:

Use two stacks:

```1. Push root to first stack.
2. Loop while first stack is not empty
2.1 Pop a node from first stack and push it to second stack(res[])
2.2 Push left and right children of the popped node to first stack
3. Pop all elements from second stack(res.reverse())```

Solution:

``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var postorderTraversal = function(root) {
let res = []; // use res as another stack
// empty tree case:
if (root === null) return res;
let stack = [];

// postorder visit: left, right, root
stack.push(root);
while (stack.length !== 0) {
let cur = stack.pop();
// treat it as stack, store them in reverse order:
// i.e root, left, right
res.push(cur.val);
if (cur.left) stack.push(cur.left);
if (cur.right) stack.push(cur.right);
}

// we can pop all elements one by one, or just reverse them
return res.reverse();
};``````