# Diagonal Traverse LeetCode Solution

Here, We see Diagonal Traverse LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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]

## Diagonal Traverse Leetcode SolutionC++

class Solution {
public:
vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
vector<int> res;
map<int, vector<int>> mp;
for(int i = 0 ; i < mat.size() ; i++)
for(int j = 0 ; j < mat[0].size() ; j++)
mp[i + j].push_back(mat[i][j]);
for(auto i : mp) {
if((i.first)%2 == 0)
reverse(i.second.begin(), i.second.end());
for(auto k : i.second) res.push_back(k);
}
return res;
}
};Code language: PHP (php)

## Diagonal Traverse Leetcode SolutionJava

class Solution {
public int[] findDiagonalOrder(int[][] mat) {
if (mat == null) {
throw new IllegalArgumentException("Input matrix is null");
}
if (mat.length == 0 || mat[0].length == 0) {
return new int[0];
}
int rows = mat.length;
int cols = mat[0].length;
int[] result = new int[rows * cols];
int r = 0;
int c = 0;
for (int i = 0; i < result.length; i++) {
result[i] = mat[r][c];
if ((r + c) % 2 == 0) {
if (c == cols - 1) {
r++;
} else if (r == 0) {
c++;
} else {
r--;
c++;
}
} else {
if (r == rows - 1) {
c++;
} else if (c == 0) {
r++;
} else {
r++;
c--;
}
}
}
return result;
}
}Code language: PHP (php)

## Diagonal Traverse SolutionJavaScript

var findDiagonalOrder = function(mat) {
let rows = mat.length;
let cols = mat[0].length;
let result = new Array(rows + cols - 1).fill(null).map(() => []);
for(let row = 0; row < rows; row++) {
for(let col = 0; col < cols; col++) {
if((row + col) % 2 === 0) result[row + col].unshift(mat[row][col]);
else result[row + col].push(mat[row][col]);
}
}
return result.flat();
};Code language: JavaScript (javascript)

## Diagonal Traverse SolutionPython

class Solution(object):
def findDiagonalOrder(self, mat):
d={}
for i in range(len(mat)):
for j in range(len(mat[i])):
if i + j not in d:
d[i+j] = [mat[i][j]]
else:
d[i+j].append(mat[i][j])
ans= []
for entry in d.items():
if entry[0] % 2 == 0:
[ans.append(x) for x in entry[1][::-1]]
else:
[ans.append(x) for x in entry[1]]
return ans
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