二叉树-二叉搜索树中第K小的元素

二叉搜索树中第K小的元素

给定一个二叉搜索树的根节点 root ，和一个整数 k ，请你设计一个算法查找其中第 k 小的元素（从 1 开始计数）。

输入：根结点，整型变量
输出：二叉树结点的值
思路：使用中序遍历，遍历到的第k个数就是第k小的元素

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int res = 0;
    int count = 0;
    public int kthSmallest(TreeNode root, int k) {
        inorder(root, k);
        return res;
    }
    public void inorder(TreeNode root, int k){
        if(root == null){
            return;
        }
        inorder(root.left, k);
        if(k == count){
            return;
        }
        if(++count == k){
            res = root.val;
        }
        inorder(root.right, k);
    }
}

if(k == count) return;
加入此行代码的原因是当遍历到第k个元素时，可以停止遍历，不用再加入其他的结点

---

二刷使用迭代法

class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Deque<TreeNode> stack = new LinkedList<>();
        while(root != null || !stack.isEmpty()){
            while(root != null){
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            k--;
            if(k == 0){
                break;
            }
            root = root.right;
        }
        return root.val;
    }
}

使用优先队列维护

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b-a);
        Deque<TreeNode> stack = new LinkedList<>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.poll();
            if(pq.size() < k){
                pq.add(node.val);
            }else if(pq.peek() > node.val){
                pq.poll();
                pq.add(node.val);
            }
            if(node.left != null){
                stack.push(node.left);
            }
            if(node.right != null){
                stack.push(node.right);
            }
        }
        return pq.peek();
    }
}