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
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public List<List<Integer>> levelOrderBottom(TreeNode root) { 12 List<List<Integer>> res = new ArrayList<List<Integer>>(); 13 if (root == null) return res; 14 Queue<TreeNode> queue = new LinkedList<TreeNode>(); 15 queue.offer(root); 16 while (!queue.isEmpty()) { 17 List<Integer> item = new ArrayList<Integer>(); 18 int size = queue.size(); 19 for (int i=0; i<size; i++) { 20 TreeNode cur = queue.poll(); 21 item.add(cur.val); 22 if (cur.left != null) { 23 queue.add(cur.left); 24 } 25 if (cur.right != null) { 26 queue.add(cur.right); 27 } 28 } 29 res.add(0, new ArrayList<Integer>(item)); 30 } 31 return res; 32 } 33 }
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)
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { 12 ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>> (); 13 if (root == null) return lists; 14 LinkedList<TreeNode> queue = new LinkedList<TreeNode>(); 15 queue.add(root); 16 int ParentNumInQ = 1; 17 int ChildNumInQ = 0; 18 ArrayList<Integer> list = new ArrayList<Integer>(); 19 while (!queue.isEmpty()) { 20 TreeNode cur = queue.poll(); 21 list.add(cur.val); 22 ParentNumInQ--; 23 if (cur.left != null) { 24 queue.add(cur.left); 25 ChildNumInQ++; 26 } 27 if (cur.right != null) { 28 queue.add(cur.right); 29 ChildNumInQ++; 30 } 31 if (ParentNumInQ == 0) { 32 ParentNumInQ = ChildNumInQ; 33 ChildNumInQ = 0; 34 lists.add(list); 35 list = new ArrayList<Integer>(); 36 } 37 } 38 Collections.reverse(lists); 39 return lists; 40 } 41 }
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:
1 public class Solution { 2 public List<List<Integer>> levelOrderBottom(TreeNode root) { 3 List<List<Integer>> wrapList = new LinkedList<List<Integer>>(); 4 levelMaker(wrapList, root, 0); 5 return wrapList; 6 } 7 8 public void levelMaker(List<List<Integer>> list, TreeNode root, int level) { 9 if(root == null) return; 10 if(level >= list.size()) { 11 list.add(0, new LinkedList<Integer>()); 12 } 13 levelMaker(list, root.left, level+1); 14 levelMaker(list, root.right, level+1); 15 list.get(list.size()-level-1).add(root.val); 16 } 17 }