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?

 

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

 //后序遍历：left->right->root
public class Solution {                  //方法与前两题先序、中序遍历一模一样
    
     public ArrayList<Integer> postorderTraversal(ArrayList<Integer> list,TreeNode root){
        if(root==null) return list;
        if(root.left!=null) postorderTraversal(list,root.left);
        if(root.right!=null) postorderTraversal(list,root.right);
        list.add(root.val);
        return list;
    }
    
    public List<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if(root==null) return result;
        
        if(root.left!=null) postorderTraversal(result,root.left);
        if(root.right!=null) postorderTraversal(result,root.right);
        result.add(root.val);
        
        return result;
    }
}

 //利用栈的迭代写法，思想参考中序遍历注释，只是while重新压入栈的顺序变化
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();      
        if(root==null) return result;
        
        Stack<TreeNode> nodeStack = new Stack<TreeNode>();
        Stack<Integer> countStack = new Stack<Integer>();
        nodeStack.push(root);
        countStack.push(0);
        
        while(!nodeStack.empty()){
            TreeNode node = nodeStack.pop();
            int count = countStack.pop();
            if(count==1){
                result.add(node.val);
            }else{
                nodeStack.push(node);
                countStack.push(1);
                if(node.right!=null){
                    nodeStack.push(node.right);
                    countStack.push(0);
                }
                if(node.left!=null){
                    nodeStack.push(node.left);
                    countStack.push(0);
                }
            }
        }

        return result;      
    }
}