Given input matrix = [ [ 5, 1, 9,11], [ 2, 4, 8,10], [13, 3, 6, 7], [15,14,12,16] ], rotate the input matrix in-place such that it becomes: [ [15,13, 2, 5], [14, 3, 4, 1], [12, 6, 8, 9], [16, 7,10,11] ]

Solution: 2 Passes

First pass: mirror around diagonal
Second pass: mirror around y axis

Time complexity: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

void rotate(vector<vector<int>>& matrix) {

const int n = matrix.size();

for (int i = 0; i < n; ++i)

for (int j = i + 1; j < n; ++j)

swap(matrix[i][j], matrix[j][i]);

for (int i = 0; i < n; ++i)

reverse(begin(matrix[i]), end(matrix[i]));

}

};

Problem

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

But the following [1,2,2,null,3,null,3] is not:

Note:
Bonus points if you could solve it both recursively and iteratively.

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

C++

class Solution {

public:

bool isSymmetric(TreeNode* root) {

return isMirror(root, root);

}

private:

bool isMirror(TreeNode* root1, TreeNode* root2) {

if (!root1 && !root2) return true;

if (!root1 || !root2) return false;

return root1->val == root2->val

&& isMirror(root1->right, root2->left)

&& isMirror(root1->left, root2->right);

}

};

Python

"""

Author: Huahua

Running time: 48 ms

"""

class Solution:

def isSymmetric(self, root):

def isMirror(root1, root2):

if not root1 and not root2: return True

if not root1 or not root2: return False

return root1.val == root2.val \

and isMirror(root1.left, root2.right) \

and isMirror(root2.left, root1.right)

return isMirror(root, root)

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Solution: 2 Passes

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Problem

Solution: Recursion

C++

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