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]
]

思路：先序遍历

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();

        if (root==null){
            return res;
        }

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.add(root);

        int curL = 0;
        while(!queue.isEmpty()){
            List<Integer> temp = new ArrayList<Integer>();
            curL = queue.size();
            for (int i=0;i<curL;i++){
                TreeNode peek = queue.remove();
                temp.add(peek.val);
                if (peek.left!=null){
                    queue.add(peek.left);
                }
                if (peek.right!=null){
                    queue.add(peek.right);
                }
            }
            res.add(temp);
        }
        return res;
    }
}

LG版code

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();

        if (root == null) return result;
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);

        while(!q.isEmpty()){
            int size = q.size();
            ArrayList<Integer> intList = new ArrayList<>();

            for (int i = 0; i<size; i++){
                TreeNode tn = q.poll();
                intList.add(tn.val);

                if (tn.left != null) {
                    q.add(tn.left);
                }
                if (tn.right != null) {
                    q.add(tn.right);
                }
        }
        result.add(intList);
    }
    return result;
    }
}