[LeetCode][Java] Recover Binary Search Tree
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题目:
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?题意:
二叉搜索树中有两个元素被错误的交换了
在不改变二叉搜索树结构的前提下,恢复原来的二叉搜索树
注意:
能否只用常数空间来完成呢?
算法分析:
* 例如1,4,3,2,5,6
* 1.当我们读到4的时候,发现是正序的,不做处理
* 2.但是遇到3时,发现逆序,将4存为第一个错误节点,3存为第二个错误节点
* 3.继续往后,发现3,2又是逆序了,那么将第2个错误节点更新为2
* 如果是这样的序列:1,4,3,5,6同上,得到逆序的两个节点为4和3。
* 需要调换的是第一个逆序的第一个元素和第二个逆序的第二个元素
* 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。
AC代码:
<span style="font-family:Microsoft YaHei;font-size:12px;">public class Solution { TreeNode first = null,second = null,pre = null; public void recoverTree(TreeNode root) { findNode(root); swap(first, second); } private void findNode(TreeNode root) { if(root == null) { return; } if(root.left != null ) findNode(root.left); if(pre != null && root.val < pre.val) { if(first == null) { first = pre; } second = root; } pre = root; if(root.right != null) findNode(root.right); } private void swap(TreeNode node1,TreeNode node2) { int tem = node1.val; node1.val = node2.val; node2.val = tem; }}</span>
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