Recover Binary Search Tree leetcode java
2014-08-05 02:54
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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?
题解:
解决方法是利用中序遍历找顺序不对的两个点,最后swap一下就好。
因为这中间的错误是两个点进行了交换,所以就是大的跑前面来了,小的跑后面去了。
所以在中序便利时,遇见的第一个顺序为抵减的两个node,大的那个肯定就是要被recovery的其中之一,要记录。
另外一个,要遍历完整棵树,记录最后一个逆序的node。
简单而言,第一个逆序点要记录,最后一个逆序点要记录,最后swap一下。
因为Inorder用了递归来解决,所以为了能存储这两个逆序点,这里用了全局变量,用其他引用型遍历解决也可以。
代码如下:
1 public class Solution {
2 TreeNode pre;
3 TreeNode first;
4 TreeNode second;
5
6 public void inorder(TreeNode root){
7 if(root == null)
8 return;
9
inorder(root.left);
if(pre == null){
pre = root; //pre指针初始
}else{
if(pre.val > root.val){
if(first == null){
first = pre;//第一个逆序点
}
second = root; //不断寻找最后一个逆序点
}
pre = root; //pre指针每次后移一位
}
inorder(root.right);
}
public void recoverTree(TreeNode root) {
pre = null;
first = null;
second = null;
inorder(root);
if(first!=null && second!=null){
int tmp = first.val;
first.val = second.val;
second.val = tmp;
}
}
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?
题解:
解决方法是利用中序遍历找顺序不对的两个点,最后swap一下就好。
因为这中间的错误是两个点进行了交换,所以就是大的跑前面来了,小的跑后面去了。
所以在中序便利时,遇见的第一个顺序为抵减的两个node,大的那个肯定就是要被recovery的其中之一,要记录。
另外一个,要遍历完整棵树,记录最后一个逆序的node。
简单而言,第一个逆序点要记录,最后一个逆序点要记录,最后swap一下。
因为Inorder用了递归来解决,所以为了能存储这两个逆序点,这里用了全局变量,用其他引用型遍历解决也可以。
代码如下:
1 public class Solution {
2 TreeNode pre;
3 TreeNode first;
4 TreeNode second;
5
6 public void inorder(TreeNode root){
7 if(root == null)
8 return;
9
inorder(root.left);
if(pre == null){
pre = root; //pre指针初始
}else{
if(pre.val > root.val){
if(first == null){
first = pre;//第一个逆序点
}
second = root; //不断寻找最后一个逆序点
}
pre = root; //pre指针每次后移一位
}
inorder(root.right);
}
public void recoverTree(TreeNode root) {
pre = null;
first = null;
second = null;
inorder(root);
if(first!=null && second!=null){
int tmp = first.val;
first.val = second.val;
second.val = tmp;
}
}
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