938. Range Sum of BST

Given the root node of a binary search tree, return the sum of values of all nodes with a value in the range [low, high].

Example 1:

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

Example 2:

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

Constraints:

  • The number of nodes in the tree is in the range [1, 2 * 104].
  • 1 <= Node.val <= 105
  • 1 <= low <= high <= 105
  • All Node.val are unique.
class Solution {
    int res = 0;
    public int rangeSumBST(TreeNode root, int L, int R) {
        dfs(root, L, R);
        return res;
    }
    public void dfs(TreeNode root, int L, int R) {
        if(root == null) return;
        if(root.val >= L && root.val <= R) res += root.val;
        if(L < root.val) dfs(root.left, L, R);
        if(R > root.val) dfs(root.right, L, R);
    }
}

Bst:左小于root,右大于root。判断每个点是不是在区间,然后判断往左还是往右。

原文地址:https://www.cnblogs.com/wentiliangkaihua/p/13984909.html