[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 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);

}

}

};

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