102. Binary Tree Level Order Traversal

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]
]

M1: BFS

建立一个queue用于存放每一层的节点，先把根节点放进queue。在一个循环里，把queue中的值全部pop出来，存成一个数组，再分别找他们的子节点，放入queue中。直到queue为空时结束。

time: O(n), space: O(2^h)　　-- 最多一层节点个数

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        Queue<TreeNode> q = new LinkedList<>();
        
        q.offer(root);
        while(!q.isEmpty()) {
            List<Integer> level = new ArrayList<>();
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode tmp = q.poll();
                level.add(tmp.val);
                
                if(tmp.left != null) {
                    q.offer(tmp.left);
                }
                if(tmp.right != null) {
                    q.offer(tmp.right);
                }
            }
            res.add(level);
        }
        return res;
    }
}

M2: recursion

用一个变量level表示层数，当递归到上一层（level=res.size)，新建一个新的list存数

time: O(n), space: O(height)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        levelOrder(root, 0, res);
        return res;
    }
    
    public void levelOrder(TreeNode root, int level, List<List<Integer>> res) {
        if(root == null) {
            return;
        }
        if(res.size() == level) {
            res.add(new ArrayList<>());
        }
        res.get(level).add(root.val);
        
        levelOrder(root.left, level + 1, res);
        levelOrder(root.right, level + 1, res);
    }
}