**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]

.

牛解法：postorder就是L-R-ROOT，这里先ROOT-R-L，再把结果reverse

public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> results = new ArrayList<Integer>();
Deque<TreeNode> stack = new ArrayDeque<TreeNode>();
while (!stack.isEmpty() || root != null) {
if (root != null) {
stack.push(root);
results.add(root.val);
root = root.right;
} else {
root = stack.pop().left;
}
}
Collections.reverse(results);
return results;
}
}

关于deque：https://docs.oracle.com/javase/7/docs/api/java/util/Deque.html

reference：https://leetcode.com/discuss/9736/accepted-code-with-explaination-does-anyone-have-better-idea