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]
]

Tree类题目。

##Approach 1: DFS 可采用DFS的思想来解。

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

##Approach 2: BFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public 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> curtLevel = new ArrayList<>();
            for(int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                curtLevel.add(node.val);
                if(node.left != null) queue.offer(node.left);
                if(node.right != null) queue.offer(node.right);
            }
            res.add(curtLevel);
        }
        return res;
    }
}