二叉查找树中搜索区间

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给定两个值 k1 和 k2（k1 < k2）和一个二叉查找树的根节点。找到树中所有值在 k1 到 k2 范围内的节点。即打印所有x (k1 <= x <= k2) 其中 x 是二叉查找树的中的节点值。返回所有升序的节点值。

如果有 k1 = 10 和 k2 = 22, 你的程序应该返回 [12, 20, 22].

    20   /  \  8   22 / \4   12

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /**     * @param root: The root of the binary search tree.     * @param k1 and k2: range k1 to k2.     * @return: Return all keys that k1<=key<=k2 in ascending order.     */    vector<int> searchRange(TreeNode* root, int k1, int k2) {        // write your code here        vector<int> res;        helper(root,k1,k2,res);        return res;    }    void helper(TreeNode* root,int k1,int k2,vector<int>&res){        if(root==NULL) return;        if(root->val<=k2&&root->val>=k1){            helper(root->left,k1,k2,res);            res.push_back(root->val);            helper(root->right,k1,k2,res);        }         else if(root->val>k2) helper(root->left,k1,k2,res);        else helper(root->right,k1,k2,res);    }};

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