[LintCode] Invert Binary Tree 翻转二叉树

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上的原题，可参见我之前的博客Invert Binary Tree 翻转二叉树。

解法一：

// Recursion
class Solution {
public:
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    void invertBinaryTree(TreeNode *root) {
        if (!root) return;
        TreeNode *tmp = root->left;
        root->left = root->right;
        root->right = tmp;
        invertBinaryTree(root->left);
        invertBinaryTree(root->right);
    }
};

解法二：

// Non-Recursion
class Solution {
public:
    /**
     * @param root: a TreeNode, the root of the binary tree
     * @return: nothing
     */
    void invertBinaryTree(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);
        }
    }
};