小黑昨晚论文又写了1000字，干到了一点，实习任务遇到了一些挑战的leetcode之旅:剑指 Offer II 052. 展平二叉搜索树&剑指 Offer II 054. 所有大于等于节点的值之和

剑指 Offer II 052. 展平二叉搜索树

小黑代码

# 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 increasingBST(self, root: TreeNode) -> TreeNode:
        # 初始化中序前结点和开始结点
        self.pre = None
        self.start =None
        # 递归函数
        def dfs(node):
            if not node:
                return
            dfs(node.left)
            # 结点操作
            if self.pre:
                self.pre.right = node
                self.pre.left = None
            else:
                self.start = node
            self.pre = node
            dfs(node.right)
        dfs(root)
        if self.pre:
            self.pre.left = None
        return self.start

递归法

# 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 increasingBST(self, root: TreeNode) -> TreeNode:
        # 初始化头结点
        head = TreeNode()
        self.link = head
        
        def dfs(node):
            if not node:
                return
            dfs(node.left)
            # 结点操作
            self.link.right = node
            self.link = node
            self.link.left = None
            dfs(node.right)
        dfs(root)
        return head.right

剑指 Offer II 054. 所有大于等于节点的值之和

小黑代码

# 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 convertBST(self, root: TreeNode) -> TreeNode:
        self.sum = 0
        # 逆向中序递归函数
        def dfs(node):
            if not node:
                return
            dfs(node.right)
            # 不断迭代sum
            self.sum += node.val
            node.val = self.sum
            dfs(node.left)
        dfs(root)
        return root

Mirris中序

# 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 convertBST(self, root: TreeNode) -> TreeNode:
        sum_ = 0
        # Mirris中序
        node = root
        while node:
            if node.right:
                # 找到其最左结点
                curr = node.right
                while curr.left and curr.left != node:
                    curr = curr.left
                # node的右子树是否遍历完成
                if curr.left:
                    # 结点操作
                    sum_ += node.val
                    node.val = sum_
                    curr.left = None
                    node = node.left
                else:
                    curr.left = node
                    node = node.right
            else:
                sum_ += node.val
                node.val = sum_
                node = node.left
        return root

小黑生活

买的筋膜枪终于到货了

昨天写论文写到一点，早上没起来吃早饭，在麦当劳吃的，希望这段日子快点儿过去，毕业与工作顺利

尽管很疲惫，依然坚持英语与运动

华为OD算法笔试过啦，但是还是由于学历问题转不了正，希望小黑未来几年的努力规避学历问题！！

吉野家牛肉饭午餐

咖啡甜点算法，无意间发现坐在我对面的小姐姐好漂亮～

开启动态规划之旅啦～～～～～～～～