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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。

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

先遍历右子树，访问根节点，再遍历左子树。遍历到第 k 个结点时，存储结果。

Python3

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

class Solution:
    t, res = 0, -1
    def kthLargest(self, root: TreeNode, k: int) -> int:
        self.t = k
        self._traverse(root)
        return self.res

    def _traverse(self, node):
        if node:
            self._traverse(node.right)
            self.t -= 1
            if self.t == 0:
                self.res = node.val
                return
            self._traverse(node.left)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int t;
    private int res;
    public int kthLargest(TreeNode root, int k) {
        t = k;
        traverse(root);
        return res;
    }

    private void traverse(TreeNode node) {
        if (node != null) {
            traverse(node.right);
            --t;
            if (t == 0) {
                res = node.val;
                return;
            }
            traverse(node.left);
        }
    }
}

JavaScript

/**
 * 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 res
    let t = 0
    function traversal(node) {
        if(!node) return
        traversal(node.right)
        if(++t === k) res = node.val
        traversal(node.left)
    }
    traversal(root)
    return res
};

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