# Diagonal Traverse – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Given an `m x n`

matrix `mat`

, return *an array of all the elements of the array in a diagonal order*.

**Example 1:**

Input:mat = [[1,2,3],[4,5,6],[7,8,9]]Output:[1,2,4,7,5,3,6,8,9]

**Example 2:**

Input:mat = [[1,2],[3,4]]Output:[1,2,3,4]

**Constraints:**

`m == mat.length`

`n == mat[i].length`

`1 <= m, n <= 10`

^{4}`1 <= m * n <= 10`

^{4}`-10`

^{5}<= mat[i][j] <= 10^{5}

# C++ Diagonal Traverse LeetCode Solution

```
``````
vector<int> findDiagonalOrder(vector<vector<int>>& matrix) {
if(matrix.empty()) return {};
const int N = matrix.size();
const int M = matrix[0].size();
vector<int> res;
for(int s = 0; s <= N + M - 2; ++s)
{
// for all i + j = s
for(int x = 0; x <= s; ++x)
{
int i = x;
int j = s - i;
if(s % 2 == 0) swap(i, j);
if(i >= N || j >= M) continue;
res.push_back(matrix[i][j]);
}
}
return res;
}
```

# Java Diagonal Traverse LeetCode Solution

```
``````
public int[] findDiagonalOrder(int[][] matrix) {
if (matrix.length == 0) return new int[0];
int r = 0, c = 0, m = matrix.length, n = matrix[0].length, arr[] = new int[m * n];
for (int i = 0; i < arr.length; i++) {
arr[i] = matrix[r][c];
if ((r + c) % 2 == 0) { // moving up
if (c == n - 1) { r++; }
else if (r == 0) { c++; }
else { r--; c++; }
} else { // moving down
if (r == m - 1) { c++; }
else if (c == 0) { r++; }
else { r++; c--; }
}
}
return arr;
}
```

# Python 3 Diagonal Traverse LeetCode Solution

```
``````
class Solution(object):
def findDiagonalOrder(self, matrix):
"""
:type matrix: List[List[int]]
:rtype: List[int]
"""
d={}
#loop through matrix
for i in range(len(matrix)):
for j in range(len(matrix[i])):
#if no entry in dictionary for sum of indices aka the diagonal, create one
if i + j not in d:
d[i+j] = [matrix[i][j]]
else:
#If you've already passed over this diagonal, keep adding elements to it!
d[i+j].append(matrix[i][j])
# we're done with the pass, let's build our answer array
ans= []
#look at the diagonal and each diagonal's elements
for entry in d.items():
#each entry looks like (diagonal level (sum of indices), [elem1, elem2, elem3, ...])
#snake time, look at the diagonal level
if entry[0] % 2 == 0:
#Here we append in reverse order because its an even numbered level/diagonal.
[ans.append(x) for x in entry[1][::-1]]
else:
[ans.append(x) for x in entry[1]]
return ans
```

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