Given a non-empty binary search tree and a target value, find k values in the BST that are closest to the target.
Note:
Given target value is a floating point.
You may assume k is always valid, that is: k ≤ total nodes.
You are guaranteed to have only one unique set of k values in the BST that are closest to the target.
Follow up:
Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?
Solution：two Stack
思路： in order 正反two pass, 分別put至stack，再從stacks求得結(jié)果
Time Complexity: O(N) Space Complexity: O(N)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> closestKValues(TreeNode root, double target, int k) {
List<Integer> res = new ArrayList<>();
Stack<Integer> s1 = new Stack<>(); // predecessors
Stack<Integer> s2 = new Stack<>(); // successors
inorder(root, target, false, s1);
inorder(root, target, true, s2);
while (k-- > 0) {
if (s1.isEmpty()) {
res.add(s2.pop());
}
else if (s2.isEmpty()) {
res.add(s1.pop());
}
else if (Math.abs(s1.peek() - target) < Math.abs(s2.peek() - target)) {
res.add(s1.pop());
}
else {
res.add(s2.pop());
}
}
return res;
}
// inorder traversal
private void inorder(TreeNode root, double target, boolean reverse, Stack<Integer> stack) {
if (root == null) return;
inorder(reverse ? root.right : root.left, target, reverse, stack);
// early terminate, no need to traverse the whole tree
if ((reverse && root.val <= target) || (!reverse && root.val > target)) return;
// track the value of current node
stack.push(root.val);
inorder(reverse ? root.left : root.right, target, reverse, stack);
}
}
