**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