中序遍历2345678
代码:
class Solution: # 返回对应节点TreeNode def KthNode(self, pRoot, k): # write code here self.res = [] self.mid(pRoot) return self.res[k-1] if 0<k<=len(self.res) else None def mid(self, root): if not root: return None self.mid(root.left) self.res.append(root) self.mid(root.right)参考资料