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99. Recover Binary Search Tree
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Example 1:
Input: [1,3,null,null,2] 1 / 3 \ 2 Output: [3,1,null,null,2] 3 / 1 \ 2
Example 2:
Input: [3,1,4,null,null,2] 3 / \ 1 4 / 2 Output: [2,1,4,null,null,3] 2 / \ 1 4 / 3
Follow up:
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A solution using O(n) space is pretty straight forward.
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Could you devise a constant space solution?
思路分析
二叉搜索树中序遍历时，就是一个升序排列的序列，在遍历过程中，就可以检查序列的大小。要求使用 O(1) 的空间复杂度，那么只能使用 Morris 遍历。
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一刷
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/**
* 使用 Morris 遍历找出错误节点，然后交换其值
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-06-23 22:21:59
*/
public void recoverTree(TreeNode root) {
if (root == null) return;
TreeNode curr = root;
TreeNode mostRight = null;
TreeNode[] errorNodes = new TreeNode[2];
TreeNode prior = null;
while (curr != null) {
mostRight = curr.left;
if (mostRight != null) {
while (mostRight.right != null && mostRight.right != curr) {
mostRight = mostRight.right;
}
if (mostRight.right == null) {
mostRight.right = curr;
curr = curr.left;
continue;
} else {
mostRight.right = null;
}
}
if (prior != null) {
if (prior.val > curr.val) {
if (errorNodes[0] == null) {
errorNodes[0] = curr;
errorNodes[1] = prior;
} else {
errorNodes[0] = curr;
}
}
}
prior = curr;
curr = curr.right;
}
int temp = errorNodes[0].val;
errorNodes[0].val = errorNodes[1].val;
errorNodes[1].val = temp;
}