05 August 2008
Given a binary search tree, write a function 1
kthSmallest
Note:
You may assume k is always valid, 1 ≤ k ≤ BST’s total elements.
Example 1:
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Input: root = [3,1,4,null,2], k = 1
3
/ \
1 4
\
2
Output: 1
Example 2:
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Input: root = [5,3,6,2,4,null,null,1], k = 3
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/ \
3 6
/ \
2 4
/
1
Output: 3
Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
使用中序遍历，遍历到第k个元素就是第k小的，作为结果返回，因为BST的中序遍历的结果刚好是从小到大排列，同样，也可以通过递归和迭代来做。
迭代解法，将k左子树的元素放入到stack中，如果不够k再放入右子树的元素，然后到k后就返回当前值
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int kthSmallest(TreeNode root, int k) {
Stack<TreeNode> stack = new Stack<>();
TreeNode current = root;
while(current != null || !stack.isEmpty()) {
while (current != null) {
stack.push(current);
current = current.left;
}
current = stack.pop();
k--;
if (k == 0) {
return current.val;
}
current = current.right; // here no need to push to stack
}
throw new IllegalArgumentException("There is no kth smallest element.");
}
}
中序遍历，先递归左子树，当前结点做计算和判断，不够再递归右子树，到k个时返回即可
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
int result = 0;
int count = 0;
public int kthSmallest(TreeNode root, int k) {
count = k;
inOrder(root);
return result;
}
private void inOrder(TreeNode node) {
if (node == null) {
return;
}
inOrder(node.left); //中序先traverse先到leftmost
count--;
if (count == 0) {
result = node.val;
return;
}
inOrder(node.right);// 不夠再traverse右sub tree
}
}