Binary Tree Postorder Traversal

Postorder: first, visit left child, then, parent, last, is to visit right child.

The postorder traversal result of above tree is {4,6,5,2,3,1}.

Key different here is that we print right child before we print parent node. Therefore, we need a mark for parent node. Only when its left child and right child are both printed, it can be printed out.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class MarkTreeNode {
    TreeNode node;
    int mark;
    MarkTreeNode(TreeNode n, int x) { node = n; mark = x; }
} 

public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> path = new ArrayList<Integer>();
        Deque<MarkTreeNode> stack = new ArrayDeque<MarkTreeNode>();
        TreeNode p = root;
        MarkTreeNode m = null;
        while (p != null || !stack.isEmpty()) {
            while (p != null) {
                m = new MarkTreeNode(p, 1);
                stack.push(m);
                p = p.left;
            }
            if (!stack.isEmpty()) {
                m = stack.peek();
                if (m.mark == 1) {
                    p = m.node.right;
                    m.mark = 2;
                } else { // m.mark == 2
                    stack.pop();
                    path.add(m.node.val);
                }
            }
        }
        return path;
    }
}