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 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
这道题又是一道2sum的变种题,博主一直强调,平生不识TwoSum,刷尽LeetCode也枉然!只要是两数之和的题,一定要记得用哈希表来做,这道题只不过是把数组变成了一棵二叉树而已,换汤不换药,我们遍历二叉树就行,然后用一个哈希set,在递归函数函数中,如果node为空,返回false。如果k减去当前结点值在哈希set中存在,直接返回true;否则就将当前结点值加入哈希set,然后对左右子结点分别调用递归函数并且或起来返回即可,参见代码如下:
解法一:
public: bool findTarget(TreeNode* root, int k) { if (!root) return false; unordered_set<int> s; return helper(root, k, s); } bool helper(TreeNode* node, int k, unordered_set<int>& s) { if (!node) return false; if (s.count(k - node->val)) return true; s.insert(node->val); return helper(node->left, k, s) || helper(node->right, k, s); } };
我们也可以用层序遍历来做,这样就是迭代的写法了,但是利用哈希表的精髓还是没变的,参见代码如下:
解法二:
public: bool findTarget(TreeNode* root, int k) { if (!root) return false; unordered_set<int> s; queue<TreeNode*> q{{root}}; while (!q.empty()) { auto t = q.front(); q.pop(); if (s.count(k - t->val)) return true; s.insert(t->val); if (t->left) q.push(t->left); if (t->right) q.push(t->right); } return false; } };
参考资料:
https://discuss.leetcode.com/topic/98440/java-c-three-simple-methods-choose-one-you-like
https://discuss.leetcode.com/topic/100110/my-c-non-recursive-solution-using-unordered_set-and-stack
本文转自博客园Grandyang的博客,原文链接:[LeetCode] Two Sum IV - Input is a BST 两数之和之四 - 输入是二叉搜索树
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