leetcode_99_Recover Binary Search Tree

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描述：

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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

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：

首先中序遍历二叉查找树并将遍历的节点存储到一个list中，然后对list中的值进行比较，查找出位置出现变化的两个结点，将两个结点的值进行互换，完成本题的要求。

但是呢，对于如何发现位置出现变化的两个结点是本题的重点和难点，具体判断条件可以参见下面的程序，最后对查到的结点进行取舍也是一大问题，一般符合判断条件的结点会出现三个，我取的是第一个和第三个，这不好讲清楚，具体可自行推敲。

代码：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public void recoverTree(TreeNode root) {List<TreeNode>list=new ArrayList<TreeNode>();     if(root==null) return;     Stack<TreeNode>st=new Stack<TreeNode>();     st.push(root);     TreeNode top=null;     while(!st.empty())     {     top=st.peek();     while(top.left!=null)     {     st.push(top.left);     top=top.left;     }     while(top.right==null)     {     list.add(top);     st.pop();     if(!st.empty())     top=st.peek();     else     break;     }     if(!st.empty())     {     list.add(top);     st.pop();     st.push(top.right);     }          }     int len=list.size()-1;     List<Integer>indexList=new ArrayList<Integer>();     if(list.get(0).val>list.get(1).val)     indexList.add(0);     for(int i=1;i<len;i++)     {     if(list.get(i).val<list.get(i-1).val||list.get(i).val>list.get(i+1).val)     indexList.add(i);     }     if(list.get(len).val<list.get(len-1).val)     indexList.add(len);     TreeNode node1=list.get(indexList.get(0));     TreeNode node2=list.get(indexList.get(indexList.size()-1));     int temp=node1.val;     node1.val=node2.val;     node2.val=temp;         }}

结果：

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