102. Binary Tree Level Order Traversal ------层序遍历

 

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]
]


层序遍历 利用queue

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(root == nullptr) return res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            res.push_back(vector<int>());
            int cnt = q.size();
            for(int i = 1; i <= cnt;++i) {
                TreeNode* node = q.front();q.pop();
                res.back().push_back(node->val);
                if(node->left != nullptr) q.push(node->left);
                if(node->right != nullptr) q.push(node->right);
            }
        }
        return res;
    }
};

 

 1 class Solution {
 2     public List<List<Integer>> levelOrder(TreeNode root) {
 3         List<List<Integer>> res = new ArrayList<List<Integer>>();
 4         Queue<TreeNode> queue = new LinkedList<TreeNode>();
 5         if(root==null) return res;
 6         queue.add(root);
 7         while (!queue.isEmpty()) {
 8             int levlnum = queue.size();
 9             List<Integer> res_temp = new ArrayList<Integer>();
10             for (int i = 0;i<levlnum ;i++ ){ //把每层的左右节点都保存到queue里 
11                 //并讲当层的值从queue里弹出，加到res_temp 中
12             if(queue.peek().left!=null) queue.add(queue.peek().left);
13             if(queue.peek().right!=null) queue.add(queue.peek().right);
14 
15                 res_temp.add(queue.poll().val);
16             }
17             res.add(res_temp);
18         }
19         return res;
20     }
21 }