代码随想录算法训练营第21天|669. 修剪二叉搜索树、108.将有序数组转换为二叉搜索树、538.把二叉搜索树转换为累加树

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#算法

打卡Day21

1.669. 修剪二叉搜索树
2.108.将有序数组转换为二叉搜索树
3.538.把二叉搜索树转换为累加树

1.669. 修剪二叉搜索树

题目链接：669. 修剪二叉搜索树
文档讲解：代码随想录

我的思路：这道题和昨天450.删除二叉搜索树中的节点很像，如果要删除节点，还是将删除节点的左子树移动到右子树最左面的节点上。一开始用的是前序遍历。在力扣上是不通过的，我觉得可能是中间节点处理顺序导致的，然后改成后序遍历就可以了，自底向上递归。

class Solution(object):
    def trimBST(self, root, low, high):
        """
        :type root: TreeNode
        :type low: int
        :type high: int
        :rtype: TreeNode
        """
        if not root:
            return root
        #自底向上，后序遍历
        root.left = self.trimBST(root.left, low, high)
        root.right = self.trimBST(root.right, low, high)
        
        if root.val < low or root.val > high:
            if not root.right:
                return root.left
            cur = root.right
            while cur.left:
                cur = cur.left
            cur.left = root.left
            return root.right
            
        return root

思路：不用重构二叉树，以下图为例，low = 1， high = 3，0 是要删除的节点，只需要将 0 的右孩子直接赋给3的左孩子就可以了。
在这里插入图片描述

class Solution(object):
    def trimBST(self, root, low, high):
        """
        :type root: TreeNode
        :type low: int
        :type high: int
        :rtype: TreeNode
        """
        if not root:
            return root
        if root.val < low:
            #在右子树寻找符合区间的节点
            return self.trimBST(root.right, low, high)
        if root.val > high:
            return self.trimBST(root.left, low, high)
        root.left = self.trimBST(root.left, low, high)
        root.right = self.trimBST(root.right, low, high)
        return root

迭代法：

class Solution(object):
    def trimBST(self, root, low, high):
        """
        :type root: TreeNode
        :type low: int
        :type high: int
        :rtype: TreeNode
        """
        if not root:
            return root
        
        #寻找合适的头节点
        while root and (root.val < low or root.val > high):
            if root.val < low:
                root = root.right
            else:
                root = root.left
        
        cur = root
        #处理左孩子元素小于low的情况
        while cur:
            while cur.left and cur.left.val < low:
                #删除节点
                cur.left = cur.left.right
            cur = cur.left
        
        cur = root
        while cur:
            while cur.right and cur.right.val > high:
                cur.right = cur.right.left
            cur = cur.right
        
        return root

2.108.将有序数组转换为二叉搜索树

题目链接：108.将有序数组转换为二叉搜索树
文档讲解：代码随想录

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        root = self.traversal(nums, 0, len(nums) - 1)
        return root
        
    def traversal(self, nums, left, right):
        if left > right:
            #定义的是左闭右闭区间，所以此时为空节点
            return None
        mid = left + (right - left) // 2
        root = TreeNode(nums[mid])
        root.left = self.traversal(nums, left, mid - 1)
        root.right = self.traversal(nums, mid + 1, right)
        return root

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if not nums:
            return 
        mid = len(nums) // 2
        root = TreeNode(nums[mid])
        root.left = self.sortedArrayToBST(nums[:mid])
        root.right = self.sortedArrayToBST(nums[mid + 1:])
        return root

迭代法：

class Solution(object):
    def sortedArrayToBST(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        if not nums:
            return None
        
        nodeQue = collections.deque()
        leftQue = collections.deque()
        rightQue = collections.deque()
        root = TreeNode(0)
        nodeQue.append(root)
        leftQue.append(0)
        rightQue.append(len(nums) - 1)

        while nodeQue:
            cur = nodeQue.popleft()
            left = leftQue.popleft()
            right = rightQue.popleft()
            mid = left + (right - left) // 2
            cur.val = nums[mid]

            #处理左区间
            if left < mid:
                cur.left = TreeNode(0)
                nodeQue.append(cur.left)
                leftQue.append(left)
                rightQue.append(mid - 1)
            if right > mid:
                cur.right = TreeNode(0)
                nodeQue.append(cur.right)
                leftQue.append(mid + 1)
                rightQue.append(right)
        return root

3.538.把二叉搜索树转换为累加树

题目链接：538.把二叉搜索树转换为累加树
文档讲解：代码随想录

思路：二叉搜索树是有序的，从树中可以看出累加顺序为右中左，然后顺序累加。本题需要利用pre指针来记录当前遍历节点cur的前一个节点。

class Solution(object):
    def convertBST(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        self.pre = 0
        #self.count = 0
        self.traversal(root)
        return root
    
    def traversal(self, cur):
        if not cur:
            return
        self.traversal(cur.right)
        cur.val += self.pre
        self.pre = cur.val
        # self.count += cur.val
        # cur.val = self.count
        self.traversal(cur.left)

迭代法：

class Solution(object):
    def convertBST(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root:
            return root
        stack = []
        cur = root
        pre = 0
        while cur or stack:
            if cur:
                stack.append(cur)
                cur = cur.right
            else:
                cur = stack.pop()
                cur.val += pre
                pre = cur.val
                cur = cur.left
        return root