[Lintcode] Invert binary tree flips two forks

Given n points on a 2D plane, find the maximum number of points that lie on the same straight line.

Example

Given 4 points:,,, (1,2) (3,6) (0,0) (1,3) .

The maximum number is 3 .

Leecode on the original topic, see my previous blog Invert binary tree flipping two fork trees.

Solution One:

//recursionclassSolution { Public:    /** * @param root:a TreeNode, the root of the binary tree * @return: Nothing*/    voidInvertbinarytree (TreeNode *root) {        if(!root)return; TreeNode*tmp = root->Left ; Root->left = root->Right ; Root->right =tmp; Invertbinarytree (Root-Left ); Invertbinarytree (Root-Right ); }};

Solution Two:

//non-recursionclassSolution { Public:    /** * @param root:a TreeNode, the root of the binary tree * @return: Nothing*/    voidInvertbinarytree (TreeNode *root) {        if(!root)return; Queue<TreeNode*>Q;        Q.push (root);  while(!Q.empty ()) {TreeNode* node =Q.front (); Q.pop (); TreeNode*tmp = node->Left ; Node->left = node->Right ; Node->right =tmp; if(node->left) Q.push (node->Left ); if(node->right) Q.push (node->Right ); }    }};

[Lintcode] Invert binary tree flips two forks