145.Binary Tree Postorder Traversal

题目：
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].

解答：
最主要的思想是先存root的话，整个存储的顺序会变反，所以要插入存储进去。
1.Iterative解答：

public class Solution {public List<Integer> postorderTraversal(TreeNode root) {List<Integer> result = new ArrayList<Integer>();Stack<TreeNode> stack = new Stack<TreeNode>();if (root == null) return result;stack.push(root);while (!stack.isEmpty()) {TreeNode node = stack.pop();result.add(0, node.val);if (node.left != null) stack.push(node.left);if (node.right != null) stack.push(node.right);}return result;}
}

2.Recursive解答：

//Recursive
public void Helper(TreeNode root, List<Integer> result) {if (root == null) return;result.add(0, root.val);Helper(root.right, result);Helper(root.left, result);
}public List<Integer> postorderTraversal(TreeNode root) {List<Integer> result = new ArrayList<Integer>();Helper(root, result);return result;
}

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