653 Two Sum IV - Input is a BST 两数之和 IV - 输入 BST
Description:
Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.
Example:
Example 1:
Input:
5
/ \
3 6
/ \ \
2 4 7
Target = 9
Output: True
Example 2:
Input:
5
/ \
3 6
/ \ \
2 4 7
Target = 28
Output: False
题目描述:
给定一个二叉搜索树和一个目标结果，如果 BST 中存在两个元素且它们的和等于给定的目标结果，则返回 true。
示例 :
案例 1:
输入:
5
/ \
3 6
/ \ \
2 4 7
Target = 9
输出: True
案例 2:
输入:
5
/ \
3 6
/ \ \
2 4 7
Target = 28
输出: False
思路:
- 遍历 BST, 将结点的值存入 set中, 每次判断元素是否在 set内即可, 可以通过剪枝加快搜索速度
- 根据 BST的特性, 将其中序遍历转化成升序数组用双指针判断
时间复杂度O(n), 空间复杂度O(n)
代码:
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:
bool findTarget(TreeNode* root, int k)
{
in_order(root);
int i = 0, j = list.size() - 1;
while (i < j)
{
if (list[i] + list[j] == k) return true;
else if (list[i] + list[j] > k) j--;
else if (list[i] + list[j] < k) i++;
}
return false;
}
private:
vector<int> list;
void in_order(TreeNode* root)
{
if (root)
{
in_order(root -> left);
list.push_back(root -> val);
in_order(root -> right);
}
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean findTarget(TreeNode root, int k) {
Set<Integer> set = new HashSet<>();
return preOrder(root, set, k);
}
private boolean preOrder(TreeNode root, Set<Integer> set, int k) {
if (root == null) return false;
if (set.contains(k - root.val)) return true;
set.add(root.val);
return preOrder(root.left, set, k) || preOrder(root.right, set, k);
}
}
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 findTarget(self, root: TreeNode, k: int) -> bool:
d, result = {}, False
def search(root):
nonlocal result
if not root or result:
return
if root.val in d:
result = True
d[k - root.val] = True
search(root.left)
search(root.right)
search(root)
return result
