【LeetCode】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]
]

二叉树的层序遍历，这道题还要求分别输出每一层。层序遍历需要用一个队列来存储根节点，然后找到根节点的左右子树，将根节点pop出来，剩下的就是下一层的所有节点，用for循环遍历，输出每一层的值。直接上代码吧，代码描述的更清楚一些。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        queue<TreeNode*> que;
        que.push(root);
        if(root==NULL) return res;
        while(!que.empty()){
            vector<int> oneLevel;
            int size = que.size();
            for(int i=0;i<size;i++){
                TreeNode* tr = que.front();
                que.pop();
                oneLevel.push_back(tr->val);
                if(tr->left) que.push(tr->left);
                if(tr->right) que.push(tr->right);
            }
            res.push_back(oneLevel);
        }
        return res;
    }
};