Leetcode: Binary Tree Level Order Transversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},
    3
   / \
  9  20
    /  \
   15   7
return its bottom-up level order traversal as:
[
  [15,7],
  [9,20],
  [3]
]

第二遍方法：

这道题在groupon面经里面有，有一个follow up 是能不能右对齐输出。那就在29行记录每一行的最大size，然后在输出的时候根据最大size补齐空格

 /**
  * Definition for binary tree
  * 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<Integer>> res = new ArrayList<List<Integer>>();
         if (root == null) return res;
         Queue<TreeNode> queue = new LinkedList<TreeNode>();
         queue.offer(root);
         while (!queue.isEmpty()) {
             List<Integer> item = new ArrayList<Integer>();
             int size = queue.size();
             for (int i=0; i<size; i++) {
                 TreeNode cur = queue.poll();
                 item.add(cur.val);
                 if (cur.left != null) {
                     queue.add(cur.left);
                 }
                 if (cur.right != null) {
                     queue.add(cur.right);
                 }
             }
             res.add(0, new ArrayList<Integer>(item));
         }
         return res;
     }
 }

在Binary Tree Level Order Transversal的基础上难度：20，只需要对最后结果做一个倒序就好。格式是Collections.reverse(List<?> list)

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
         ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>> ();
         if (root == null) return lists;
         LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
         queue.add(root);
         int ParentNumInQ = 1;
         int ChildNumInQ = 0;
         ArrayList<Integer> list = new ArrayList<Integer>();
         while (!queue.isEmpty()) {
             TreeNode cur = queue.poll();
             list.add(cur.val);
             ParentNumInQ--;
             if (cur.left != null) {
                 queue.add(cur.left);
                 ChildNumInQ++;
             }
             if (cur.right != null) {
                 queue.add(cur.right);
                 ChildNumInQ++;
             }
             if (ParentNumInQ == 0) {
                 ParentNumInQ = ChildNumInQ;
                 ChildNumInQ = 0;
                 lists.add(list);
                 list = new ArrayList<Integer>();
             }
         }
         Collections.reverse(lists);
         return lists;
     }
 }

注意38行的写法，Collections.reverse()跟Arrays.sort()函数一样，都是void返回型，然后改变作用在argument上

还有if (root == null) return null;

ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();这样会出错：

Input:{}Output:nullExpected:[]

DFS 做法：

 public class Solution {
         public List<List<Integer>> levelOrderBottom(TreeNode root) {
             List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
             levelMaker(wrapList, root, 0);
             return wrapList;
         }

         public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {
             if(root == null) return;
             if(level >= list.size()) {
                 list.add(0, new LinkedList<Integer>());
             }
             levelMaker(list, root.left, level+1);
             levelMaker(list, root.right, level+1);
             list.get(list.size()-level-1).add(root.val);
         }
     }