145. Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

Follow up: Recursive solution is trivial, could you do it iteratively?

这道题和94,144题是一个系列，刚好是三种遍历方式。

94：https://blog.****.net/qq_39638957/article/details/89372757

144：https://blog.****.net/qq_39638957/article/details/89377357

145这道题是一个hard难度的。利用LinkedList的双向特性，可以转换为和preorder一样的题目

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这里画图对这三道题目进行总结：

代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        LinkedList<Integer> res = new LinkedList<Integer>();//双向链表
        ArrayDeque<TreeNode> stack = new ArrayDeque<TreeNode>();
        TreeNode cur = root;
        while(cur!=null || !stack.isEmpty()){
            if(cur!=null){
                stack.push(cur);
                res.addFirst(cur.val);
                cur = cur.right;
            }
            else{
                cur = stack.pop().left;
            }
        }
        return res;
    }
}