【LeetCode】102.Binary Tree Level Order Traversal(Medium)解题报告

【LeetCode】102.Binary Tree Level Order Traversal(Medium)解题报告

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

  Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

Solution1:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 二叉树的层序遍历，很经常出现的一个问题
 第一种方法：queue
 time : O(n)
 space : O(n)
 */
class Solution {
    public List<List<Integer>> levelOrder(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(list);
        }
        return res;
    }
}

Solution2:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 二叉树的层序遍历，很经常出现的一个问题
 第二种方法：先序遍历求层序遍历
 time : O(n)
 space : O(n)
 */
class Solution {
    public List<List<Integer>> levelOrder(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(new ArrayList<>());
        }
        res.get(level).add(root.val);
        helper(res,root.left,level+1);
        helper(res,root.right,level+1);
    }
}

Date：2018年3月14日