Search Range in Binary Search Tree

Description

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.

Example
If k1 = 10 and k2 = 22, 
then your function should return [12, 20, 22].
    20
   /  \
  8   22
 / \
4   12

Link

Lintcode_ladder

Method

1. DFS, in-order traversal
2. x

Example

1. 1

/**
 * 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.
     */
    // iteration
    vector<int> searchRange(TreeNode* root, int k1, int k2) {
        // write your code here
        if (!root) {
            return {};
        }
        vector<int> result;
        stack<TreeNode*> stk;
        TreeNode* cur = root;
        while (cur || !stk.empty()) {
            while (cur) {
                stk.push(cur);
                cur = cur->left;
            }
            cur = stk.top();
            stk.pop();
            if (cur->val <= k2 && cur->val >= k1) {
                result.push_back(cur->val);
            }
            cur = cur->right;
        }
        return result;
    }
    // ----------------------------------------------------------------
    // recursion
    vector<int> searchRange(TreeNode* root, int k1, int k2) {
        if (!root) {
            return {};
        }
        vector<int> result;
        helper(result, root, k1, k2);
        return result;
    }
    void helper(vector<int>& result, TreeNode* root, int k1, int k2) {
        if (!root) {
            return;
        }
        // root->val > k1, search left
        if (root->val > k1) {
            helper(result, root->left, k1, k2);
        }
        // match
        if (root->val <= k2 && root->val >= k1) {
            result.push_back(root->val);
        }
        // root->val < k2, search right
        if (root->val < k2) {
            helper(result, root->right, k1 ,k2);
        }
        return;
    }
};

Similar problems

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