问题描述:题意就是二叉树中有两个节点交换了,恢复结构。
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
算法分析:其实还是利用中序遍历,每次记录前一个访问的节点,和当前节点就行比较,如果逆序则记录下来。两个节点交换有两种情况,一种是相邻节点交换,那么只有一个逆序,非相邻节点交换,则有两个逆序。
public class RecoverBinarySearchTree
{
TreeNode mistake1 = null;
TreeNode mistake2 = null;
TreeNode pre = null;
public void recoverTree(TreeNode root)
{
inorderTraverse(root);
int temp = mistake1.val;
mistake1.val = mistake2.val;
mistake2.val = temp;
}
public void inorderTraverse(TreeNode root)
{
if(root == null)
{
return;
}
inorderTraverse(root.left);
if(pre != null)
{
if(pre.val >= root.val)//将当前要访问的根节点和pre比较
{
if(mistake1 == null)
{
mistake1 = pre;
}
mistake2 = root;
}
}
pre = root;//访问完根节点时,将根节点置为pre,相当于pre指针后移一位
inorderTraverse(root.right);
}
}