十一第三题Binary Tree Level Order Traversal II

public class Solution {    public List<List<Integer>> levelOrderBottom(TreeNode root) {        List<List<Integer>> list = new ArrayList<List<Integer>>();        if (root == null)            return list;        List<TreeNode> mainList = new ArrayList<TreeNode>();        List<Integer> tmpList = new ArrayList<Integer>();        int curr = 0;        Stack<List<Integer>> stack = new  Stack<List<Integer>>();        mainList.add(root);         tmpList.add(root.val);        while (mainList.size() > curr) {            int tmp = mainList.size();            List<Integer> newTmpList = new ArrayList<Integer>();            for (int i = curr; i < tmp; i++)                newTmpList.add(tmpList.get(i));            stack.push(newTmpList);            while (tmp > curr) {                TreeNode node = mainList.get(curr++);                if (node.left != null) {                    mainList.add(node.left);                    tmpList.add(node.left.val);                }                                    if (node.right != null) {                    mainList.add(node.right);                    tmpList.add(node.right.val);                }                                }        }        while (!stack.empty())            list.add(stack.pop());                return list;    }}

思路很简单，利用层次遍历，然后利用stack先进后出的思想。