LeetCode：102. Binary Tree Level Order Traversal（二叉树的层次遍历）

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]
]

---

方法1：深度优先（递归算法）

/**
 * 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<List<Integer>>();
        levelHelper(res, root, 0);
        return res;
    }
    
    public void levelHelper(List<List<Integer>> res, TreeNode root, int height) {
        if (root == null) return;
        if (height >= res.size()) {
            res.add(new LinkedList<Integer>());
        }
        res.get(height).add(root.val);
        levelHelper(res, root.left, height+1);
        levelHelper(res, root.right, height+1);
    }
}

时间复杂度：O(n)

空间复杂度：O(n)

---

方法2：一种非递归的方式，利用栈的形式

/**
 * 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<List<Integer>>();
        Queue<TreeNode> queue=new LinkedList<>();
        queue.add(root);
        while(!queue.isEmpty()){
            List<Integer> list=new ArrayList();
            int size=queue.size();
            for(int i=0;i<size;i++){
                TreeNode p=queue.poll();
                list.add(p.val);
                if(p.left!=null){
                    queue.add(p.left);
                }
                if(p.right!=null){
                    queue.add(p.right);
                }
            }
             res.add(list);
        }       
        return res;
    }
}

时间复杂度：O(n)

空间复杂度：O(n)