题目
给定一棵二叉搜索树,请找出其中第k大的节点。
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
代码
Python
# 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:
# 采用DFS中的中序遍历的倒序,即递减序列,“右,根,左”
self.res, self.k = None, k # 返回值,k
def dfs(root):
if not root: return # 父节点为空,终止
dfs(root.right) # 递归遍历右儿子
self.k-=1
if self.k==0: self.res=root.val # 返回二叉搜索树的第k大节点,终止遍历
if self.k>0: dfs(root.left) # 递归遍历左儿子
dfs(root)
return self.res
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 res;
int kthLargest(TreeNode* root, int k) {
dfs(root,k);
return res;
}
void dfs(TreeNode* root, int &k) {
if (!root) return;
dfs(root->right,k);
k--;
if (k==0) res=root->val;
if (k>0) dfs(root->left,k);
}
};