Class RecoverBinarySearchTree

src/tree/binary/traverse.rs:107–107

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? 示例 1： 输入：root = [1,3,null,null,2] 输出：[3,1,null,null,2] 解释：3 不能是 1 左孩子，因为 3 > 1 。交换 1 和 3 使二叉搜索树有效。 示例 2： 输入：root = [3,1,4,null,null,2] 输出：[2,1,4,null,null,3] 解释：2 不能在 3 的右子树中，因

/// 输出：[2,1,4,null,null,3]
/// 解释：2 不能在 3 的右子树中，因为 2 < 3 。交换 2 和 3 使二叉搜索树有效。
pub struct RecoverBinarySearchTree;

/// Given two binary trees, write a function to check if they are equal or not.
/// Two binary trees are considered equal if they are structurally identical and