99. Recover Binary Search Tree
You are given the root of a binary search tree (BST), where the values of exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.
Example 1:
Input: root = [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.
Example 2:
Input: root = [3,1,4,null,null,2]
Output: [2,1,4,null,null,3]
Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.
Constraints:
- The number of nodes in the tree is in the range [2, 1000].
- -231 <= Node.val <= 231 - 1
Follow up: A solution using O(n) space is pretty straight-forward. Could you devise a constant O(1) space solution?
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private TreeNode first;
private TreeNode second;
private TreeNode pre;
public void recoverTree(TreeNode root) {
if(root==null) return;
first = null;
second = null;
pre = null;
inorder(root);
int temp = first.val;
first.val = second.val;
second.val = temp;
}
private void inorder(TreeNode root){
if(root==null) return;
inorder(root.left);
if(first==null && (pre==null ||pre.val>=root.val)){
first = pre;
}
if(first!=null && pre.val>=root.val){
second = root;
}
pre = root;
inorder(root.right);
}
}将节点指针放在数组里，寻找两个逆向的位置，先从前往后找到第一个逆序的位置，然后从后往前找到第二个逆序的位置，交换两个指针的值