二叉树的三种遍历（递归实现）

/**
 * struct TreeNode {
 *  int val;
 *  struct TreeNode *left;
 *  struct TreeNode *right;
 * };
 */

class Solution {
  public:
    /**
     *
     * @param root TreeNode类 the root of binary tree
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > threeOrders(TreeNode* root) {
        vector<vector<int>> vv;
        vector<int> one = preOrder(root);
        vector<int> two = midOrder(root);
        vector<int> three = postOrder(root);
        vv.push_back(one);
        vv.push_back(two);
        vv.push_back(three);
        return vv;
    }

    vector<int> preOrder(TreeNode* root) {

        vector<int> v;

        if (root == NULL) return v;

        v.push_back(root->val);

        vector<int> lv = preOrder(root->left);
        convert(lv, v);
        vector<int> rv = preOrder(root->right);
        convert(rv, v);

        return v;
    }


    vector<int> midOrder(TreeNode* root) {

        vector<int> v;
        if (root == NULL) return v;
        vector<int> lv = midOrder(root->left);
        convert(lv, v);
        v.push_back(root->val);
        vector<int> rv = midOrder(root->right);
        convert(rv, v);
        return v;
    }


    vector<int> postOrder(TreeNode* root) {

        vector<int> v;
        if (root == NULL) return v;
        vector<int> lv = postOrder(root->left);
        convert(lv, v);

        vector<int> rv = postOrder(root->right);
        convert(rv, v);

        v.push_back(root->val);

        return v;
    }


    void convert(vector<int>& a, vector<int>& b) {
        for (int i = 0; i < a.size(); i++) {
            b.push_back(a[i]);
        }
    }

};