Algorithms—107.Binary Tree Level Order Traversal II

来源：互联网发布：知乎2016 编辑：程序博客网时间：2024/05/20 19:29

思路：抄102题，循环载入每层的node，逆向赋值即可。

/** * 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<List<Integer>> levelOrderBottom(TreeNode root) {List<List<TreeNode>> answerList = new ArrayList<List<TreeNode>>();List<List<Integer>> list = new ArrayList<List<Integer>>();if (root == null) {return list;}List<TreeNode> initList = new ArrayList<TreeNode>();initList.add(root);answerList.add(initList);for (int i = 0; i < answerList.size(); i++) {List<TreeNode> newList = new ArrayList<TreeNode>();for (int j = 0; j < answerList.get(i).size(); j++) {TreeNode tree = answerList.get(i).get(j);if (tree.left != null) {newList.add(tree.left);}if (tree.right != null) {newList.add(tree.right);}}if (newList.size() != 0) {answerList.add(newList);}}for (int i = answerList.size()-1; i >=0; i--) {List<Integer> l = new ArrayList<Integer>();for (int j = 0; j < answerList.get(i).size(); j++) {l.add(answerList.get(i).get(j).val);}list.add(l);}return list;}}

耗时：304ms，中上游

0 0