Question

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

方法一

# 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)

/**
 * 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);
  }
}

/**
 * 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);
  }
};

/**
 * 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)
}

/**
 * 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);
}