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剑指 Offer II 056. 二叉搜索树中两个节点之和

题目描述

给定一个二叉搜索树的 根节点 root 和一个整数 k , 请判断该二叉搜索树中是否存在两个节点它们的值之和等于 k 。假设二叉搜索树中节点的值均唯一。

 

示例 1：

输入: root = [8,6,10,5,7,9,11], k = 12
输出: true
解释: 节点 5 和节点 7 之和等于 12

示例 2：

输入: root = [8,6,10,5,7,9,11], k = 22
输出: false
解释: 不存在两个节点值之和为 22 的节点

 

提示：

  • 二叉树的节点个数的范围是  [1, 104].
  • -104 <= Node.val <= 104
  • root 为二叉搜索树
  • -105 <= k <= 105

 

注意：本题与主站 653 题相同： https://leetcode.cn/problems/two-sum-iv-input-is-a-bst/

解法

用哈希表记录访问过的节点。

Python3

# 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 findTarget(self, root: TreeNode, k: int) -> bool:
        def find(root):
            if not root:
                return False
            if k - root.val in nodes:
                return True
            nodes.add(root.val)
            return find(root.left) or find(root.right)

        nodes = set()
        return find(root)

Java

/**
 * 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 {
    private Set<Integer> nodes;

    public boolean findTarget(TreeNode root, int k) {
        nodes = new HashSet<>();
        return find(root, k);
    }

    private boolean find(TreeNode root, int k) {
        if (root == null) {
            return false;
        }
        if (nodes.contains(k - root.val)) {
            return true;
        }
        nodes.add(root.val);
        return find(root.left, k) || find(root.right, k);
    }
}

TypeScript

/**
 * 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 findTarget(root: TreeNode | null, k: number): boolean {
    let nodes: Set<number> = new Set();
    return find(root, k, nodes);
}

function find(root: TreeNode | null, k: number, nodes: Set<number>): boolean {
    if (!root) return false;
    if (nodes.has(k - root.val)) return true;
    nodes.add(root.val);
    return find(root.left, k, nodes) || find(root.right, k, nodes);
}

C++

/**
 * 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:
    unordered_set<int> nodes;

    bool findTarget(TreeNode* root, int k) {
        return find(root, k);
    }

    bool find(TreeNode* root, int k) {
        if (!root) return false;
        if (nodes.count(k - root->val)) return true;
        nodes.insert(root->val);
        return find(root->left, k) || find(root->right, k);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func findTarget(root *TreeNode, k int) bool {
	nodes := make(map[int]bool)

	var find func(root *TreeNode, k int) bool
	find = func(root *TreeNode, k int) bool {
		if root == nil {
			return false
		}
		if nodes[k-root.Val] {
			return true
		}
		nodes[root.Val] = true
		return find(root.Left, k) || find(root.Right, k)
	}
	return find(root, k)
}

...