leetcode 1145. Binary Tree Postorder Traversal ----- java

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?

求后序遍历，要求不使用递归。

使用栈，从后向前添加。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List list = new ArrayList<Integer>();

        if( root == null )
            return list;
        Stack<TreeNode> stack = new Stack<TreeNode>();

        stack.push(root);

        while( !stack.isEmpty() ){

            TreeNode node = stack.pop();
            list.add(0,node.val);
            if( node.left != null )
                stack.push(node.left);
            if( node.right != null )
                stack.push(node.right);

        }
        return list;

    }
}