LeetCode 938. Range Sum of BST

原题链接在这里：https://leetcode.com/problems/range-sum-of-bst/

题目：

Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).

The binary search tree is guaranteed to have unique values.

Example 1:

Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
Output: 32

Example 2:

Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
Output: 23

Note:

The number of nodes in the tree is at most 10000.
The final answer is guaranteed to be less than 2^31.

题解：

If current node is null, return 0.

If node value < L, return range sum from its right.

If node value > R, return range sum from its left.

If it is withing [L, R], return range sum from both left and right + node.val.

Time Complexity: O(n).

Space: O(logn).

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public int rangeSumBST(TreeNode root, int L, int R) {
12         if(root == null){
13             return 0;
14         }
15         
16         if(root.val < L){
17             return rangeSumBST(root.right, L, R);
18         }
19         
20         if(root.val > R){
21             return rangeSumBST(root.left, L, R);
22         }
23         
24         return rangeSumBST(root.right, L, R) + rangeSumBST(root.left, L, R) + root.val;
25     }
26 }

AC JavaScript:

 1 /**
 2  * Definition for a binary tree node.
 3  * function TreeNode(val) {
 4  *     this.val = val;
 5  *     this.left = this.right = null;
 6  * }
 7  */
 8 /**
 9  * @param {TreeNode} root
10  * @param {number} L
11  * @param {number} R
12  * @return {number}
13  */
14 var rangeSumBST = function(root, L, R) {
15     if(!root){
16         return 0;
17     }
18     
19     if(root.val < L){
20         return rangeSumBST(root.right, L, R);
21     }
22     
23     if(root.val > R){
24         return rangeSumBST(root.left, L, R);
25     }
26     
27     return root.val + rangeSumBST(root.left, L, R) + rangeSumBST(root.right, L, R);
28 };

Iteration with stack.

Time Complexity: O(n).

Space: O(logn).

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public int rangeSumBST(TreeNode root, int L, int R) {
12         if(root == null){
13             return 0;
14         }
15         
16         int sum = 0;
17         Stack<TreeNode> stk = new Stack<>();
18         stk.push(root);
19 
20         while(!stk.isEmpty()){
21             TreeNode cur = stk.pop();
22             if(cur == null){
23                 continue;
24             }
25             
26             if(cur.val < L){
27                 stk.push(cur.right);
28             }else if(cur.val > R){
29                 stk.push(cur.left);
30             }else{
31                 stk.push(cur.left);
32                 stk.push(cur.right);
33                 sum += cur.val;
34             }
35         }
36         
37         return sum;
38     }
39 }