标签：etc ret ++ middle 有序数组 find bst tar false
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
思路：
中序遍历二叉树，将节点值保存在数组里面。然后用从头和尾部遍历刚刚得到的有序数组。
void middle( TreeNode* root, vector< int >& a ) { if( root == NULL ) return; middle( root->left, a ); a.push_back( root->val ); middle( root->right, a ); } bool findTarget(TreeNode* root, int k) { vector< int > a; middle( root, a ); int n = a.size(); for( int i = 0, j = n-1; i < j; ) { if( a[i] + a[j] < k ) i++; else if( a[i] + a[j] > k ) j--; else return true; } return false; }
[leetcode-653-Two Sum IV - Input is a BST]
标签：etc ret ++ middle 有序数组 find bst tar false
原文地址：http://www.cnblogs.com/hellowooorld/p/7294151.html
