python3 解法, 执行用时: 36 ms, 内存消耗: 14.9 MB, 提交时间: 2022-08-22 15:54:48

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def printTree(self, root: Optional[TreeNode]) -> List[List[str]]:
        def calDepth(root: Optional[TreeNode]) -> int:
            h = -1
            q = [root]
            while q:
                h += 1
                tmp = q
                q = []
                for node in tmp:
                    if node.left:
                        q.append(node.left)
                    if node.right:
                        q.append(node.right)
            return h
        height = calDepth(root)

        m = height + 1
        n = 2 ** m - 1
        ans = [[''] * n for _ in range(m)]
        q = deque([(root, 0, (n - 1) // 2)])
        while q:
            node, r, c = q.popleft()
            ans[r][c] = str(node.val)
            if node.left:
                q.append((node.left, r + 1, c - 2 ** (height - r - 1)))
            if node.right:
                q.append((node.right, r + 1, c + 2 ** (height - r - 1)))
        return ans

python3 解法, 执行用时: 44 ms, 内存消耗: 14.8 MB, 提交时间: 2022-08-22 15:54:26

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def printTree(self, root: Optional[TreeNode]) -> List[List[str]]:
        def calDepth(node: Optional[TreeNode]) -> int:
            return max(calDepth(node.left) + 1 if node.left else 0, calDepth(node.right) + 1 if node.right else 0)
        height = calDepth(root)

        m = height + 1
        n = 2 ** m - 1
        ans = [[''] * n for _ in range(m)]
        def dfs(node: Optional[TreeNode], r: int, c: int) -> None:
            ans[r][c] = str(node.val)
            if node.left:
                dfs(node.left, r + 1, c - 2 ** (height - r - 1))
            if node.right:
                dfs(node.right, r + 1, c + 2 ** (height - r - 1))
        dfs(root, 0, (n - 1) // 2)
        return ans