面试题 54. 二叉搜索树的第 k 大节点

题目描述

给定一棵二叉搜索树，请找出其中第 k 大的节点的值。

 

示例 1:

输入: root = [3,1,4,null,2], k = 1
   3
  / \
 1   4
  \
   2
输出: 4

示例 2:

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

 

限制：

• 1 ≤ k ≤ 二叉搜索树元素个数

解法

方法一：反序中序遍历

由于二叉搜索树的中序遍历是升序的，因此可以反序中序遍历，即先递归遍历右子树，再访问根节点，最后递归遍历左子树。

这样就可以得到一个降序的序列，第 $k$ 个节点就是第 $k$ 大的节点。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉搜索树的节点个数。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def kthLargest(self, root: TreeNode, k: int) -> int:
        def dfs(root):
            nonlocal k, ans
            if root is None or k == 0:
                return
            dfs(root.right)
            k -= 1
            if k == 0:
                ans = root.val
            dfs(root.left)

        ans = 0
        dfs(root)
        return ans

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int k;
    private int ans;

    public int kthLargest(TreeNode root, int k) {
        this.k = k;
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null || k == 0) {
            return;
        }
        dfs(root.right);
        if (--k == 0) {
            ans = root.val;
        }
        dfs(root.left);
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int kthLargest(TreeNode* root, int k) {
        int ans = 0;
        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root || !k) {
                return;
            }
            dfs(root->right);
            if (--k == 0) {
                ans = root->val;
            }
            dfs(root->left);
        };
        dfs(root);
        return ans;
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func kthLargest(root *TreeNode, k int) (ans int) {
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil || k == 0 {
			return
		}
		dfs(root.Right)
		k--
		if k == 0 {
			ans = root.Val
		}
		dfs(root.Left)
	}
	dfs(root)
	return
}

/**
 * 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 kthLargest(root: TreeNode | null, k: number): number {
    let res = 0;
    const dfs = (root: TreeNode | null) => {
        if (root == null) {
            return;
        }
        dfs(root.right);
        if (--k === 0) {
            res = root.val;
            return;
        }
        dfs(root.left);
    };
    dfs(root);
    return res;
}

// 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::rc::Rc;
use std::cell::RefCell;

impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, arr: &mut Vec<i32>) {
        if let Some(node) = root {
            let node = node.borrow();
            Solution::dfs(&node.right, arr);
            arr.push(node.val);
            Solution::dfs(&node.left, arr)
        }
    }
    pub fn kth_largest(root: Option<Rc<RefCell<TreeNode>>>, k: i32) -> i32 {
        let mut arr = vec![];
        Solution::dfs(&root, &mut arr);
        arr[(k - 1) as usize]
    }
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} k
 * @return {number}
 */
var kthLargest = function (root, k) {
    let ans = 0;
    const dfs = root => {
        if (!root || !k) {
            return;
        }
        dfs(root.right);
        if (--k == 0) {
            ans = root.val;
        }
        dfs(root.left);
    };
    dfs(root);
    return ans;
};

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int ans;
    private int k;

    public int KthLargest(TreeNode root, int k) {
        this.k = k;
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null || k == 0) {
            return;
        }
        dfs(root.right);
        if (--k == 0) {
            ans = root.val;
        }
        dfs(root.left);
    }
}

方法二

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func kthLargest(root *TreeNode, k int) int {
	ch := make(chan int)
	ctx, cancel := context.WithCancel(context.Background())
	defer cancel()
	go inorder(ctx, root, ch)
	for ; k > 1; k-- {
		<-ch
	}
	return <-ch
}

func inorder(ctx context.Context, cur *TreeNode, ch chan<- int) {
	if cur != nil {
		inorder(ctx, cur.Right, ch)
		select {
		case ch <- cur.Val:
		case <-ctx.Done():
			return
		}
		inorder(ctx, cur.Left, ch)
	}
}