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]
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        if (root == null) return new LinkedList();
        Stack<TreeNode> s = new Stack<TreeNode>();
        LinkedList<Integer> res = new LinkedList();
        s.push(root);
        while(!s.empty()){
            TreeNode p = s.pop();
            res.addFirst(p.val);
            //左右根,push进stack是第一次逆,addFirst又逆回来,所以还是左右根
            if(p.left != null){
                s.push(p.left);
            }
            if(p.right != null){
                s.push(p.right);
            }
        }
        return res;
    }
}

LinkedList有addFirst方法,巧妙。

posted @ 2019-08-28 01:12  Schwifty  阅读(129)  评论(0编辑  收藏  举报