题目&＃xff1a;原题链接&＃xff08;简单&＃xff09;
与题目1038相同
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|
| Ans 1 (Python) | O(N)O(N)O(N) | O(1)O(1)O(1) | 76ms (87.34%) |
| Ans 2 (Python) | O(N)O(N)O(N) | O(1)O(1)O(1) | 68ms (98.52%) |
解法一&＃xff08;反序中序遍历&＃xff09;&＃xff1a;
def __init__(self):self.total &＃61; 0def convertBST(self, root: TreeNode) -> TreeNode:def helper(node):if not node:return 0helper(node.right)self.total &＃43;&＃61; node.valnode.val &＃61; self.totalhelper(node.left)helper(root)return root
解法二&＃xff08;解法一的优雅化&＃xff09;&＃xff1a;
def __init__(self):self.total &＃61; 0def convertBST(self, root: TreeNode) -> TreeNode:if root:self.convertBST(root.right)self.total &＃43;&＃61; root.valroot.val &＃61; self.totalself.convertBST(root.left)return root
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