[Leetcode][第99题][JAVA][恢复二叉搜索树][中序遍历]

【问题描述】[困难]

在这里插入图片描述

【解答思路】

1. 显示中序遍历

在这里插入图片描述

时间复杂度：O(N) 空间复杂度：O(N)

class Solution {public void recoverTree(TreeNode root) {List<Integer> nums = new ArrayList<Integer>();inorder(root, nums);int[] swapped = findTwoSwapped(nums);recover(root, 2, swapped[0], swapped[1]);}public void inorder(TreeNode root, List<Integer> nums) {if (root == null) {return;}inorder(root.left, nums);nums.add(root.val);inorder(root.right, nums);}public int[] findTwoSwapped(List<Integer> nums) {int n = nums.size();//第一次找到取前面 第二次找到取后面int x = -1, y = -1;for (int i = 0; i < n - 1; ++i) {if (nums.get(i + 1) < nums.get(i)) {y = nums.get(i + 1);if (x == -1) {x = nums.get(i);} else {break;}}}return new int[]{x, y};}public void recover(TreeNode root, int count, int x, int y) {if (root != null) {if (root.val == x || root.val == y) {root.val = root.val == x ? y : x;if (--count == 0) {return;}}recover(root.right, count, x, y);recover(root.left, count, x, y);}}
}

2. 隐式中序遍历栈

在这里插入图片描述

时间复杂度：O(N) 空间复杂度：O(H)
在这里插入图片描述

class Solution {public void recoverTree(TreeNode root) {Deque<TreeNode> stack = new ArrayDeque<TreeNode>();TreeNode x = null, y = null, pred = null;while (!stack.isEmpty() || root != null) {while (root != null) {stack.push(root);root = root.left;}root = stack.pop();if (pred != null && root.val < pred.val) {y = root;if (x == null) {x = pred;} else {break;}}pred = root;root = root.right;}swap(x, y);}public void swap(TreeNode x, TreeNode y) {int tmp = x.val;x.val = y.val;y.val = tmp;}
}

3. Morris 中序遍历

在这里插入图片描述

时间复杂度：O(N) 空间复杂度：O(1)
在这里插入图片描述

class Solution {public void recoverTree(TreeNode root) {TreeNode x = null, y = null, pred = null, predecessor = null;while (root != null) {if (root.left != null) {// predecessor 节点就是当前 root 节点向左走一步，然后一直向右走至无法走为止predecessor = root.left;while (predecessor.right != null && predecessor.right != root) {predecessor = predecessor.right;}// 让 predecessor 的右指针指向 root，继续遍历左子树if (predecessor.right == null) {predecessor.right = root;root = root.left;}// 说明左子树已经访问完了，我们需要断开链接else {if (pred != null && root.val < pred.val) {y = root;if (x == null) {x = pred;}}pred = root;predecessor.right = null;root = root.right;}}// 如果没有左孩子，则直接访问右孩子else {if (pred != null && root.val < pred.val) {y = root;if (x == null) {x = pred;}}pred = root;root = root.right;}}swap(x, y);}public void swap(TreeNode x, TreeNode y) {int tmp = x.val;x.val = y.val;y.val = tmp;}
}