标签：style blog http color strong io for ar
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?
算法思路：
思路1：求出中序遍历序列，空间复杂度O(n)，tip要求只可以开O(1)空间。
代码略
思路2：中序遍历，遇到每个点，都记录其前驱，记录当前指针cur的前一个节点pre，如果pre.val大于cur.val，表示有错序，多数情况错序有两次；如果有一次错序，说明就是相邻节点需要被交换。
1 public class Solution { 2 TreeNode first = null,second = null,pre = null; 3 public void recoverTree(TreeNode root) { 4 findNode(root); 5 swap(first, second); 6 } 7 private void findNode(TreeNode root){ 8 if(root == null){ 9 return; 10 } 11 if(root.left != null ) findNode(root.left); 12 if(pre != null && root.val < pre.val){ 13 if(first == null){ 14 first = pre; 15 } 16 second = root; 17 } 18 pre = root; 19 if(root.right != null) findNode(root.right); 20 } 21 private void swap(TreeNode node1,TreeNode node2){ 22 int tem = node1.val; 23 node1.val = node2.val; 24 node2.val = tem; 25 } 26 }
[leetcode]Recover Binary Search Tree,布布扣,bubuko.com
[leetcode]Recover Binary Search Tree
标签：style blog http color strong io for ar
原文地址：http://www.cnblogs.com/huntfor/p/3881570.html
