Binary Tree Postorder Traversal

LeetCode 145 | Difficulty: Easy

Easy

Problem Description

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]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [4,6,7,5,2,9,8,3,1]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

Input: root = [1]

Output: [1]

Constraints:

- The number of the nodes in the tree is in the range `[0, 100]`.

- `-100 <= Node.val <= 100`

Follow up: Recursive solution is trivial, could you do it iteratively?

Topics: Stack, Tree, Depth-First Search, Binary Tree

---

Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

Stack

Use a stack (LIFO) to track elements that need future processing. Process elements when a "trigger" condition is met (e.g., finding a smaller/larger element). Monotonic stack maintains elements in sorted order for next greater/smaller element problems.

When to use

Matching brackets, next greater element, evaluating expressions, backtracking history.

---

Solutions

Solution 1: C# (Best: 284 ms)

Metric	Value
Runtime	284 ms
Memory	N/A
Date	2018-05-03

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public IList<int> PostorderTraversal(TreeNode root) {
            Stack<TreeNode> s1 = new Stack<TreeNode>();
    Stack<TreeNode> s2 = new Stack<TreeNode>();
    List<int> postOrder = new List<int>();
        if(root==null) return postOrder;
    s1.Push(root);
    while (s1.Count != 0)
    {
        var popped = s1.Pop();
        s2.Push(popped);
        if (popped.left != null)
        {
            s1.Push(popped.left);
        }
        if (popped.right != null)
        {
            s1.Push(popped.right);
        }
    }
    while(s2.Count!=0)
    {
        postOrder.Add(s2.Pop().val);
    }
    
    return postOrder;
    }
}

---

Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$
Stack	$O(n)$	$O(n)$

---

Interview Tips

Key Points

• Start by clarifying edge cases: empty input, single element, all duplicates.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.
• Think about what triggers a pop: is it finding a match, or finding a smaller/larger element?