Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1    \     2    /   3

return [3,2,1].

Postorder is also a type of DFS. So we need to use a stack to help us control the visiting order of every node. Since we need to visit the root node at last, that means we would visit the root node in the scenario that traverses back from the children node. So we can separate this problem into two cases, one is to travel down and another is to travel back.

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<Integer> postorderTraversal(TreeNode root) {        ArrayList<Integer> result = new ArrayList<Integer>();        Stack buffer = new Stack<TreeNode>();        TreeNode prev = null;        if (root == null) {return result;}        buffer.push(root);        while (!buffer.isEmpty()) {        TreeNode cur = (TreeNode) buffer.peek();        // traverse down        if (prev == null || prev.left == cur || prev.right == cur) {if (cur.left != null) {buffer.push(cur.left);}else if (cur.right != null) {buffer.push(cur.right);}else {result.add(cur.val);buffer.pop();}}        // traverse back        else {// back from left        if (cur.left == prev) {// check right        if (cur.right != null) {buffer.push(cur.right);}        else {result.add(cur.val);buffer.pop();}}        //back from right        else {result.add(cur.val);buffer.pop();}}        prev = cur;        }        return result;    }}