Binary Tree Postorder Traversal

Author: gouthampradhan

README.md

@@ -325,6 +325,7 @@ My accepted leetcode solutions to some of the common interview problems.
 - [Closest Leaf in a Binary Tree](problems/src/tree/ClosestLeafInABinaryTree.java) (Medium)
 - [Maximum Width of Binary Tree](problems/src/tree/MaximumWidthOfBinaryTree.java) (Medium)
 - [Recover Binary Search Tree](problems/src/tree/RecoverBinarySearchTree.java) (Hard)
+- [Binary Tree Postorder Traversal](problems/src/tree/BinaryTreePostorderTraversal.java) (Hard)
 
 #### [Two Pointers](problems/src/two_pointers)
 

problems/src/tree/BinaryTreePostorderTraversal.java

@@ -0,0 +1,62 @@
+package tree;
+import java.util.*;
+/**
+ * Created by gouthamvidyapradhan on 28/07/2018.
+ * 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: O(N). Maintain a stack, for every node which you pop from stack add it to result list, push left
+ and right node to stack. Reverse the result list and return this as the answer.
+ */
+public class BinaryTreePostorderTraversal {
+    public static class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+
+    public static void main(String[] args) throws Exception{
+        TreeNode root = new TreeNode(1);
+        root.right = new TreeNode(2);
+        root.right.left = new TreeNode(3);
+        List<Integer> result = new BinaryTreePostorderTraversal().postorderTraversal(root);
+        result.forEach(System.out::println);
+    }
+
+    public List<Integer> postorderTraversal(TreeNode root) {
+        Stack<TreeNode> stack = new Stack<>();
+        List<Integer> result = new ArrayList<>();
+        if(root != null){
+            stack.push(root);
+        }
+        while(!stack.isEmpty()){
+            TreeNode node = stack.pop();
+            result.add(node.val);
+            if(node.left != null){
+                stack.push(node.left);
+            } if(node.right != null){
+                stack.push(node.right);
+            }
+        }
+        Collections.reverse(result);
+        return result;
+    }
+
+
+
+}