LeetCode 102

方法一：（迭代）

/**
 * 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>> v;
        if(root == NULL) return v;

        queue<TreeNode *> q;
        q.push(root);

        TreeNode *curNode = NULL;
        vector<int> a;
        while(!q.empty())
        {
            int len = q.size();
            for(size_t i = ; i < len; i++)
            {
                curNode = q.front(); q.pop();
                a.push_back(curNode->val);
                if(curNode->left != NULL) q.push(curNode->left);
                if(curNode->right != NULL) q.push(curNode->right);
            }
            v.push_back(a);
            a.clear();
        }

        return v;
    }
};

方法二：（递归）

class Solution
{
public:
    vector<vector<int>> levelOrder(TreeNode* root)
    {
       vector<vector<int>> v;
       levelOrderHelp(root, , v);
       return v;
    }

    void levelOrderHelp(TreeNode *root, int level, vector<vector<int>>& v)
    {
        if(root == NULL) return;

　　　　 // 注意
        if(level >= v.size()) v.push_back(vector<int>());

        v[level].push_back(root->val);
        levelOrderHelp(root->left, level + , v);
        levelOrderHelp(root->right, level + , v);
    }
};