跳转至

230. 二叉搜索树中第 K 小的元素

题目描述

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）。

 

示例 1：

输入：root = [3,1,4,null,2], k = 1
输出：1

示例 2：

输入：root = [5,3,6,2,4,null,null,1], k = 3
输出：3

 

 

提示：

  • 树中的节点数为 n 。
  • 1 <= k <= n <= 104
  • 0 <= Node.val <= 104

 

进阶：如果二叉搜索树经常被修改（插入/删除操作）并且你需要频繁地查找第 k 小的值，你将如何优化算法？

解法

方法一：中序遍历

由于二叉搜索树的性质，中序遍历一定能得到升序序列，因此可以采用中序遍历找出第 k 小的元素。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        stk = []
        while root or stk:
            if root:
                stk.append(root)
                root = root.left
            else:
                root = stk.pop()
                k -= 1
                if k == 0:
                    return root.val
                root = root.right

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
/**
 * 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) {
        Deque<TreeNode> stk = new ArrayDeque<>();
        while (root != null || !stk.isEmpty()) {
            if (root != null) {
                stk.push(root);
                root = root.left;
            } else {
                root = stk.pop();
                if (--k == 0) {
                    return root.val;
                }
                root = root.right;
            }
        }
        return 0;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        stack<TreeNode*> stk;
        while (root || !stk.empty()) {
            if (root) {
                stk.push(root);
                root = root->left;
            } else {
                root = stk.top();
                stk.pop();
                if (--k == 0) return root->val;
                root = root->right;
            }
        }
        return 0;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func kthSmallest(root *TreeNode, k int) int {
    stk := []*TreeNode{}
    for root != nil || len(stk) > 0 {
        if root != nil {
            stk = append(stk, root)
            root = root.Left
        } else {
            root = stk[len(stk)-1]
            stk = stk[:len(stk)-1]
            k--
            if k == 0 {
                return root.Val
            }
            root = root.Right
        }
    }
    return 0
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function kthSmallest(root: TreeNode | null, k: number): number {
    const dfs = (root: TreeNode | null) => {
        if (root == null) {
            return -1;
        }
        const { val, left, right } = root;
        const l = dfs(left);
        if (l !== -1) {
            return l;
        }
        k--;
        if (k === 0) {
            return val;
        }
        return dfs(right);
    };
    return dfs(root);
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(root: Option<Rc<RefCell<TreeNode>>>, res: &mut Vec<i32>, k: usize) {
        if let Some(node) = root {
            let mut node = node.borrow_mut();
            Self::dfs(node.left.take(), res, k);
            res.push(node.val);
            if res.len() >= k {
                return;
            }
            Self::dfs(node.right.take(), res, k);
        }
    }
    pub fn kth_smallest(root: Option<Rc<RefCell<TreeNode>>>, k: i32) -> i32 {
        let k = k as usize;
        let mut res: Vec<i32> = Vec::with_capacity(k);
        Self::dfs(root, &mut res, k);
        res[k - 1]
    }
}

方法二：预处理结点数

预处理每个结点作为根节点的子树的节点数。

这种算法可以用来优化频繁查找第 k 个树、而二叉搜索树本身不被修改的情况。

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right


class BST:
    def __init__(self, root):
        self.cnt = Counter()
        self.root = root
        self.count(root)

    def kthSmallest(self, k):
        node = self.root
        while node:
            if self.cnt[node.left] == k - 1:
                return node.val
            if self.cnt[node.left] < k - 1:
                k -= self.cnt[node.left] + 1
                node = node.right
            else:
                node = node.left
        return 0

    def count(self, root):
        if root is None:
            return 0
        n = 1 + self.count(root.left) + self.count(root.right)
        self.cnt[root] = n
        return n


class Solution:
    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        bst = BST(root)
        return bst.kthSmallest(k)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
/**
 * 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) {
        BST bst = new BST(root);
        return bst.kthSmallest(k);
    }
}

class BST {
    private TreeNode root;
    private Map<TreeNode, Integer> cnt = new HashMap<>();

    public BST(TreeNode root) {
        this.root = root;
        count(root);
    }

    public int kthSmallest(int k) {
        TreeNode node = root;
        while (node != null) {
            int v = node.left == null ? 0 : cnt.get(node.left);
            if (v == k - 1) {
                return node.val;
            }
            if (v < k - 1) {
                node = node.right;
                k -= (v + 1);
            } else {
                node = node.left;
            }
        }
        return 0;
    }

    private int count(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int n = 1 + count(root.left) + count(root.right);
        cnt.put(root, n);
        return n;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BST {
public:
    BST(TreeNode* root)
        : root(root) {
        count(root);
    }

    int kthSmallest(int k) {
        TreeNode* node = root;
        while (node) {
            int v = !node->left ? 0 : cnt[node->left];
            if (v == k - 1) return node->val;
            if (v < k - 1) {
                node = node->right;
                k -= (v + 1);
            } else
                node = node->left;
        }
        return 0;
    }

private:
    TreeNode* root;
    unordered_map<TreeNode*, int> cnt;

    int count(TreeNode* root) {
        if (!root) return 0;
        int n = 1 + count(root->left) + count(root->right);
        cnt[root] = n;
        return n;
    }
};

class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        BST bst(root);
        return bst.kthSmallest(k);
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
type BST struct {
    cnt  map[*TreeNode]int
    root *TreeNode
}

func newBST(root *TreeNode) *BST {
    var count func(*TreeNode) int
    cnt := map[*TreeNode]int{}
    count = func(root *TreeNode) int {
        if root == nil {
            return 0
        }
        n := 1 + count(root.Left) + count(root.Right)
        cnt[root] = n
        return n
    }
    count(root)
    return &BST{cnt, root}
}

func (bst *BST) kthSmallest(k int) int {
    node := bst.root
    for node != nil {
        v := 0
        if node.Left != nil {
            v = bst.cnt[node.Left]
        }
        if v == k-1 {
            return node.Val
        }
        if v < k-1 {
            k -= (v + 1)
            node = node.Right
        } else {
            node = node.Left
        }
    }
    return 0
}

func kthSmallest(root *TreeNode, k int) int {
    bst := newBST(root)
    return bst.kthSmallest(k)
}

评论