Recover Binary Search Tree

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. 

BST:二叉查找树。根节点大于左子树所有节点，根节点小于右子树所有节点。递归定义

题意

一棵BST中有两个节点交换了，恢复这颗树

思路

1.O(N)空间复杂度

中序遍历将节点放到list中，遍历list放到int[]数组中，对int[]数组进行排序，遍历list。list.get(i).val = nums[i]

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 import java.util.ArrayList;
11 import java.util.Arrays;
12 import java.util.List;
13 
14 public class Solution {
15     List<TreeNode> listOfInOrder = new ArrayList<TreeNode>();
16     
17     public void recoverTree(TreeNode root) {
18         inOrderTravel(root);
19         int nums[] = new int[listOfInOrder.size()];
20         for(int i = 0; i < nums.length; i++){
21             nums[i] = listOfInOrder.get(i).val;
22         }//for
23         
24         Arrays.sort(nums);                             //对数组升序排列
25         for(int i = 0; i < nums.length; i++)
26             listOfInOrder.get(i).val = nums[i];
27     }
28     
29     /**
30      * 中序遍历，节点放到list中
31      * @param root
32      */
33     private void inOrderTravel(TreeNode root){
34         if(root != null){
35             inOrderTravel(root.left);
36             listOfInOrder.add(root);
37             inOrderTravel(root.right);
38         }
39     }
40 }

 2.O(1)空间复杂度，中序遍历递归

 1 public class Solution
 2 {
 3     private TreeNode node1 = null;
 4     private TreeNode node2 = null;
 5     private TreeNode pre = null;
 6 
 7     public void recoverTree(TreeNode root) {
 8         if(root == null)
 9             return;
10         inOrderTravel(root);
11         
12         int temp = node1.val;
13         node1.val = node2.val;
14         node2.val = temp;
15     }
16 
17     private void inOrderTravel(TreeNode root){
18         if(root != null){
19             inOrderTravel(root.left);
20             if(pre != null && pre.val > root.val){
21                 if(node1 == null){
22                     node1 = pre;
23                     node2 = root;
24                 }
25                     
26                 node2 = root;
27             }//if
28             pre = root;
29             inOrderTravel(root.right);
30         }
31     }//inOrderTravel
32 
33     public static void main(String args[]){
34         TreeNode root = new TreeNode(1);
35         TreeNode node1 = new TreeNode(2);
36         TreeNode node2 = new TreeNode(3);
37 
38         root.right = node1;
39         node1.left = node2;
40 
41         Solution sol = new Solution();
42 
43         inOrderTravelOut(root);
44         System.out.println();
45         sol.recoverTree(root);
46         inOrderTravelOut(root);
47     }
48 
49     private static void inOrderTravelOut(TreeNode root){
50         if(root != null){
51             inOrderTravelOut(root.left);
52             System.out.print(root.val + " ");
53             inOrderTravelOut(root.right);
54         }
55     }
56 }

 

posted on 2015-02-04 16:52  luckygxf  阅读(142)  评论(0)  编辑  收藏  举报

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