【LeetCode】107.Binary Tree Level Order Traversal II(Easy)解题报告

【LeetCode】107.Binary Tree Level Order Traversal II(Easy)解题报告

题目地址：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/description/

题目描述：

  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,null,null,15,7],
3
/ \
9  20
/  \
15   7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]

Solution1:

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
还是两种方法，一种遍历，第二种前序遍历实现层序遍历
time : O(n)
space : O(n)
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()){
int size = queue.size();
List<Integer> list = new ArrayList<>();
for(int i=0 ; i<size ; i++){
TreeNode cur = queue.poll();
if(cur.left!=null) queue.offer(cur.left);
if(cur.right!=null) queue.offer(cur.right);
list.add(cur.val);
}
res.add(0,list);
}
return res;
}
}

Solution2:

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
还是两种方法，一种遍历，第二种前序遍历实现层序遍历
time : O(n)
space : O(n)
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> res = new ArrayList<>();
if(root == null) return res;
helper(res,root,0);
return res;
}
public static void helper(List<List<Integer>> res, TreeNode root,int level){
if(root==null) return;
if(level>=res.size()){
res.add(0,new ArrayList<>());
}
res.get(res.size()-level-1).add(root.val);
helper(res,root.left,level+1);
helper(res,root.right,level+1);
}
}

Date：2018年3月17日