Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.
Example 1:
Input: root = [3,1,4,null,2], k = 1 Output: 1
Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3 Output: 3
Constraints:
- The number of nodes in the tree is n.
- 1 <= k <= n <= 104
- 0 <= Node.val <= 104
Follow up: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?
1. DFS 中序递归：时间复杂度：O(N)
// better keep these two variables in a wrapper class
private static int number = 0;
private static int count = 0;
public int kthSmallest(TreeNode root, int k) {
count = k;
helper(root);
return number;
}
public void helper(TreeNode n) {
if (n.left != null) helper(n.left);
count--;
if (count == 0) {
number = n.val;
return;
}
if (n.right != null) helper(n.right);
}
2. DFS 中序迭代：
时间复杂度：O(N) 最好
public int kthSmallest(TreeNode root, int k) {
Stack st = new Stack<>();
while (root != null) {
st.push(root);
root = root.left;
}
while (k != 0) {
TreeNode n = st.pop();
k--;
if (k == 0) return n.val;
TreeNode right = n.right;
while (right != null) {
st.push(right);
right = right.left;
}
}
return -1; // never hit if k is valid
}
参考
https://leetcode.com/problems/kth-smallest-element-in-a-bst/discuss/63660/3-ways-implemented-in-JAVA-(Python)%3A-Binary-Search-in-order-iterative-and-recursive
