Leetcode: 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].

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

Use stack and two variables to store current and previous node. Use the relationship between the two node to decide whether to push or pop.

/**
 * 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> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        TreeNode pre = null;
        stack.push(root);
        while (!stack.empty()) {
            cur = stack.peek();
            if (pre == null || pre.left == cur || pre.right == cur) {
                if (cur.left != null) {
                    stack.push(cur.left);
                } else if (cur.right != null) {
                    stack.push(cur.right);
                }
            } else if (cur.left == pre) {
                if (cur.right != null) {
                    stack.push(cur.right);
                }
            } else {
                res.add(cur.val);
                stack.pop();
            }
            pre = cur;
        }
        
        return res;
    }
    
    
}

Above solution is still complicate to remember and implement. Another easier method is as follows. Similar to Binary Tree Preorder Traversal, but instead of right node, push left node first, and pop and push into another stack. That stack stores the reverse order of the result.

/**
 * 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> res = new ArrayList<Integer>();
        if (root == null) {
            return res;
        }
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        Stack<Integer> reverse = new Stack<Integer>();
        stack.push(root);
        while (!stack.empty()) {
            TreeNode curr = stack.pop();
            reverse.push(curr.val);
            if (curr.left != null) {
                stack.push(curr.left);
            }
            if (curr.right != null) {
                stack.push(curr.right);
            }
        }
        
        while (!reverse.empty()) {
            res.add(reverse.pop());
        }
        
        return res;
    }
    
    
}