本问题对应的 leetcode 原文链接：剑指 Offer 54. 二叉搜索树的第k大节点
问题描述
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
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
int k = 0;
int target = 0;
public int kthLargest(TreeNode root, int k) {
// 中序遍历：有序的序列 左 根 右 从小到大排序的
// 右 根 左 从大到小的排序
this.k = k;
right_root_left(root);
return target;
}
void right_root_left(TreeNode root){
if(root == null || k <= 0) return;
right_root_left(root.right);
// 打印当前根节点
k--;
if(k == 0){
target = root.val;
}
right_root_left(root.left);
}
}
时间复杂度：最差为 O(n)，平均为 logn
空间复杂度：最差为 O(n)，平均为 logn
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def kthLargest(self, root, k):
"""
:type root: TreeNode
:type k: int
:rtype: int
"""
def right_root_left(root):
if not root or self.k <= 0:
return
right_root_left(root.right)
# 打印当前根节点
self.k -= 1
if self.k == 0:
self.target = root.val
right_root_left(root.left)
self.target = 0
self.k = k
right_root_left(root)
return self.target
C++
/**
* 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 target = 0;
right_root_left(root, k, target);
return target;
}
void right_root_left(TreeNode* root, int &k, int &target){
if(!root || k <= 0) return;
right_root_left(root->right, k, target);
// 打印当前根节点
k--;
if(k == 0){
target = root->val;
}
right_root_left(root->left, k, target);
}
};
Go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func kthLargest(root *TreeNode, k int) int {
var (
target int
dfs func(node *TreeNode)
)
dfs = func(node *TreeNode) {
if node == nil || k <= 0 {
return
}
dfs(node.Right)
k--
if k == 0 {
target = node.Val
return
}
dfs(node.Left)
}
dfs(root)
return target
}
JS
/**
* 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) {
// 中序遍历：有序的序列 左 根 右 从小到大排序的
// 右 根 左 从大到小的排序
this.k = k;
right_root_left(root);
return target;
};
function right_root_left(root) {
if (root == null || k <= 0) return;
right_root_left(root.right);
// 打印当前根节点
k--;
if (k == 0) {
target = root.val;
}
right_root_left(root.left);
}
shuaidi