leetcode 每日一题 99. 恢复二叉搜索树

Posted 2020-12-12 nil_f

中序遍历排序

思路：

①按中序遍历树

②确定交换的元素x,y

③再次遍历树，改变对应节点的值

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 recoverTree(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        def inorder(r: TreeNode) -> List[int]:
            return inorder(r.left) + [r.val] + inorder(r.right) if r else []
        
        def find_two_swapped(nums: List[int]) -> (int, int):
            n = len(nums)
            x = y = -1
            for i in range(n - 1):
                if nums[i + 1] < nums[i]:
                    y = nums[i + 1]
                    if x == -1:     
                        x = nums[i]
                    else:           
                        break
            return x, y
        
        def recover(r: TreeNode, count: int):
            if r:
                if r.val == x or r.val == y:
                    r.val = y if r.val == x else x
                    count -= 1
                    if count == 0:
                        return      
                recover(r.left, count)
                recover(r.right, count)
            
        nums = inorder(root)
        x, y = find_two_swapped(nums)
        recover(root, 2)

迭代中序遍历

思路：

通过迭代构造中序遍历，并在一次遍历中找到要交换的节点。

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 recoverTree(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        stack = []
        x = y = pred = None
        
        while stack or root:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            if pred and root.val < pred.val:
                y = root
                if x is None:
                    x = pred 
                else:
                    break
            pred = root
            root = root.right

        x.val, y.val = y.val, x.val

递归中序遍历

思路：

迭代中序遍历可以转换为递归。

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 recoverTree(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        def find_two_swapped(root: TreeNode):
            nonlocal x, y, pred
            if root is None:
                return
            
            find_two_swapped(root.left)
            if pred and root.val < pred.val:
                y = root
                if x is None:
                    x = pred 
                else:
                    return
            pred = root
            find_two_swapped(root.right)
        
        x = y = pred = None
        find_two_swapped(root)
        x.val, y.val = y.val, x.val

线索二叉树

思路:

参考94.二叉树的中序遍历中的不改变树结构的线索二叉树。

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 recoverTree(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        x = y = predecessor = pred = None
        while root:
            if root.left:       
                predecessor = root.left
                while predecessor.right and predecessor.right != root:
                    predecessor = predecessor.right
                if predecessor.right is None:
                    predecessor.right = root
                    root = root.left
                else:
                    if pred and root.val < pred.val:
                        y = root
                        if x is None:
                            x = pred 
                    pred = root
                    predecessor.right = None
                    root = root.right
            else:
                if pred and root.val < pred.val:
                    y = root
                    if x is None:
                        x = pred 
                pred = root
                root = root.right
        x.val, y.val = y.val, x.val