Intuition

Peel the matrix like an onion: walk the top row, right column, bottom row, left column, then shrink the boundaries inward and repeat.

Approach

Track four boundaries: top, bottom, left, right.
Traverse top row left-to-right, right column top-to-bottom, bottom row right-to-left, left column bottom-to-top.
After each edge, move the corresponding boundary inward; stop when boundaries cross.

Solution

def spiralOrder(self, matrix):
    res = []
    top, bottom = 0, len(matrix) - 1
    left, right = 0, len(matrix[0]) - 1
    while top <= bottom and left <= right:
        for c in range(left, right + 1): res.append(matrix[top][c])
        top += 1
        for r in range(top, bottom + 1): res.append(matrix[r][right])
        right -= 1
        if top <= bottom:
            for c in range(right, left - 1, -1): res.append(matrix[bottom][c])
            bottom -= 1
        if left <= right:
            for r in range(bottom, top - 1, -1): res.append(matrix[r][left])
            left += 1
    return res

Complexity

Time — O(m*n): each element is appended exactly once.
Space — O(1): aside from the output, only boundary counters.

Common Pitfalls

Re-check top <= bottom and left <= right before the bottom row and left column to avoid double-visiting on non-square matrices.

Spiral Matrix

Example 1

Example 2

Constraints