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Q230KthSmallestElementinaBST.java
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Q230KthSmallestElementinaBST.java
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import util.TreeNode;
import java.util.ArrayDeque;
/**
* @author ahscuml
* @date 2019/5/17
* @time 23:14
*/
public class Q230KthSmallestElementinaBST {
/**
* 测试函数
* */
public static void main(String[] args) {
}
/**
* BST的特点就是中序遍历就是顺序的数组
* 按照中序遍历的结构来完成
* */
int count = 0;
int res = -1;
public int kthSmallest(TreeNode root, int k) {
helper(root, k);
return res;
}
void helper(TreeNode root, int k) {
if(root != null) {
kthSmallest(root.left, k);
if(++count == k) {
res = root.val;
}
kthSmallest(root.right, k);
}
}
/**
* 利用循环的方法
* */
public int kthSmallestIte(TreeNode root, int k) {
TreeNode cur = root;
ArrayDeque<TreeNode> stack = new ArrayDeque();
while(!stack.isEmpty() || cur != null) {
if(cur != null) {
stack.push(cur);
cur = cur.left;
} else {
cur = stack.pop();
if(--k == 0) {
return cur.val;
}
cur = cur.right;
}
}
return -1;
}
/**
* 另外一种,利用二分查找,分别统计这个节点的左右节点各有多少个节点,直到找到K个
* */
public int kthSmallestIII(TreeNode root, int k) {
int cnt = count(root.left);
if(k <= cnt) {
return kthSmallestIII(root.left, k);
} else if(k > cnt + 1) {
return kthSmallestIII(root.right, k - cnt - 1);
}
return root.val;
}
int count(TreeNode root) {
if(root == null) {
return 0;
}
return 1 + count(root.left) + count(root.right);
}
}