【Lintcode】011.Search Range in Binary Search Tree

题目：

Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree. Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST. Return all the keys in ascending order.

If k1 = 10 and k2 = 22, then your function should return [12, 20, 22].

题解：

Solution 1 ()

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) {

        vector<int> result;

        inOrder(result, root, k1, k2);

        return result;

    }

    void inOrder(vector<int> &result, TreeNode* root, int k1, int k2) {

        if (root == NULL) {

            return;

        }

        if (root->val > k1) {

            inOrder(result, root->left, k1, k2);

        }

        if (k1 <= root->val && root->val <= k2) {

            result.push_back(root->val);

        }

        if (root->val < k2) {

            inOrder(result, root->right, k1, k2);

        }

    }

};

巴特西

【Lintcode】011.Search Range in Binary Search Tree

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