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Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".思路：我们都知道中序遍历树的话就会是一个递增的序列，那么如果交换的节点刚好相邻的话，那么我们记录然后交换一下就行了，但是如果两个数是不相邻的话，比如说：12345，交换了1和5,->52341，那么这个序列就存在了两个逆序对，52,和41，那么我们记录第一次出现的第一个，第二次出现的第二个，也就是5和1，那么交换就行了。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeNode preNode = null;
private void cal(TreeNode root, List<TreeNode> res) {
if (root == null) return;
cal(root.left, res);
if (preNode != null && preNode.val > root.val) {
if (res.size() == 0) {
res.add(preNode);
res.add(root);
} else {
res.set(1, root);
}
}
preNode = root;
cal(root.right, res);
}
public void recoverTree(TreeNode root) {
if (root == null) return;
List<TreeNode> res = new ArrayList<TreeNode>();
cal(root, res);
if (res.size() > 0) {
int tmp = res.get(0).val;
res.get(0).val = res.get(1).val;
res.get(1).val = tmp;
}
}
}LeetCode Recover Binary Search Tree
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原文地址：http://blog.csdn.net/u011345136/article/details/45273973
