[LeetCode]145. 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].

后序遍历啦，递归方法，so easy

/** * 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) {if (root == null) {List<Integer> res = new  ArrayList<Integer>(0);return res;}List<Integer> res = new  ArrayList<Integer>();List<Integer> left = new  ArrayList<Integer>();List<Integer> right = new  ArrayList<Integer>();left = postorderTraversal(root.left);right = postorderTraversal(root.right);res.addAll(left);res.addAll(right);    res.add(root.val);return res;    }}