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整理了python version的逻辑。做了一下简化
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@ -23,7 +23,7 @@
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* 111.二叉树的最小深度
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@ -209,29 +209,28 @@ class Solution {
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# self.right = right
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class Solution:
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def __init__(self):
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self.pre = TreeNode()
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self.count = 0
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def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
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if root == None:
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return
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'''
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倒序累加替换:
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[2, 5, 13] -> [[2]+[1]+[0], [2]+[1], [2]] -> [20, 18, 13]
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'''
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self.traversal(root)
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return root
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def traversal(self, root: TreeNode) -> None:
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# 因为要遍历整棵树,所以递归函数不需要返回值
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# Base Case
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if not root:
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return None
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# 单层递归逻辑:中序遍历的反译 - 右中左
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self.traversal(root.right) # 右
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# 右
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self.convertBST(root.right)
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# 中
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# 中节点:用当前root的值加上pre的值
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root.val += self.pre.val # 中
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self.pre = root
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self.count += root.val
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self.traversal(root.left) # 左
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root.val = self.count
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# 左
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self.convertBST(root.left)
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return root
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```
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## Go
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