Leetcode 99. Recover Binary Search Tree

最新推荐文章于 2021-06-13 13:54:56 发布

原创最新推荐文章于 2021-06-13 13:54:56 发布 · 159 阅读

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这篇博客讨论了一种算法，用于恢复一个二叉搜索树，该树中有两个节点因错误被交换。算法首先遍历树并保存节点值，然后对值进行排序并重新赋值给节点，从而恢复BST的正确结构。这是一个O(n)空间复杂度的解决方案，挑战在于如何设计常数空间的解决方案。

Problem

You are given the root of a binary search tree (BST), where exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.

Follow up: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

Algorithm

Save the data in a list, sort and restore.

Code

# 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.
        """
        index = 0
        pL = []
        nL = []
        def travel(root):
            if root.left:
                travel(root.left)
            pL.append(root)
            nL.append(root.val)
            if root.right:
                travel(root.right)
        
        travel(root)
        nL.sort()
        for i in range (len(pL)):
            pL[i].val = nL[i]