Lintcode011.Search Range in Binary Search Tree

Posted 2020-09-17 Vincent丶

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].

    20
   /    8   22
 / 4   12

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);
        }
    }
};