将二叉搜索树变平衡-二叉树1382-python

没看答案，先中序遍历不平衡的BST，再重新构造平衡BST。

# 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 balanceBST(self, root: TreeNode) -> TreeNode:
        # 中序遍历BST得到的是递增数组
        tree_list = []

        def dfs(root):
            if not root:
                return
            
            dfs(root.left)
            tree_list.append(root.val)
            dfs(root.right)
        
        dfs(root)
        
        # 重构BST
        BST = TreeNode(0)

        def rebuildBST(tree_list, BST):
            if not tree_list:
                return None
            
            mid = len(tree_list) // 2
            BST = TreeNode(tree_list[mid])
            BST.left = rebuildBST(tree_list[:mid], BST.left)
            BST.right = rebuildBST(tree_list[mid+1:], BST.right)

            return BST

        return rebuildBST(tree_list, BST)