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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
 // eg:[10,5,30,-2,6,12,40,null,2,null,null,null,null,null,null,-1]
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        LinkedList<Integer> res = new LinkedList<Integer>();
        Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
        TreeNode cur = root;
        
        while( cur != null || !stack.isEmpty()){
            while(cur != null) {
                res.addFirst(cur.val);
                stack.push(cur);
                cur = cur.right;
            }
            
            cur = stack.pop();
            cur = cur.left;
            
        }
        
        return res;
    }
    
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
 // eg:[10,5,30,-2,6,12,40,null,2,null,null,null,null,null,null,-1]
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<Integer>();
        helper(root, res);
        return res;
    }
    
    public void helper(TreeNode root, List<Integer> res){
        if ( root == null) return;
        
        helper( root.left, res);
        helper( root.right, res);
        res.add( root.val);
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
 // eg:[10,5,30,-2,6,12,40,null,2,null,null,null,null,null,null,-1]
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<Integer>();
        
        TreeNode cur = new TreeNode(-1);
        cur.left = root;
        
        while( cur != null){
            if (cur.left == null){
                cur = cur.right;
            } else {
                
                TreeNode temp = cur.left;
                
                while( temp.right != null && temp.right != cur){
                    temp = temp.right;
                }
                
                if( temp.right == null){
                    temp.right = cur;
                    cur = cur.left;
                } else {
                    addReverseOrder(cur.left, temp, res);
                    temp.right = null;
                    cur = cur.right;
                }
                
            }
        
        }
        
        return res;
    }
    
    
    public void reverseOrder(TreeNode start, TreeNode end){
        if (start == end ) return;
        
        TreeNode x = start;
        TreeNode y = x.right;
        
        while(true){
            TreeNode temp = y.right;
            y.right = x;
            x = y;
            y = temp;
            if (x == end) break;
        }
        
    }
    public void addReverseOrder(TreeNode cur, TreeNode pre, List<Integer> res){
        
        reverseOrder(cur, pre);
        
        TreeNode temp = pre;
        while(true){
            res.add(temp.val);
            if(temp == cur) break;
            temp = temp.right;
        }
       
       reverseOrder(pre,cur);
    }
}

/*
1.先建立一个临时结点dummy，并令其左孩子为根结点root，将当前结点设置为dummy；
2.如果当前结点的左孩子为空，则将其右孩子作为当前结点；
3.如果当前结点的左孩子不为空，则找到其在中序遍历中的前驱结点
  3.1如果前驱结点的右孩子为空，将它的右孩子设置为当前结点，将当前结点更新为当前结点的左孩子；
  3.2如果前驱结点的右孩子为当前结点，倒序输出从当前结点的左孩子到该前驱结点这条路径上所有的结点。将前驱结点的右孩子设置为空，将当前结点更新为当前结点的右孩子。
4.重复以上过程，直到当前结点为空。
*/