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leetcode--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?

solution 1 using O(n) space:

 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 public class Solution {
11     public void recoverTree(TreeNode root) {
12          ArrayList<TreeNode> Inorder = InOrderTraversal(root);
13         int firstViolation = 0;
14         int secondViolation = 0;
15         int length = Inorder.size();
16         if(!Inorder.isEmpty()){
17             int i = 0;
18             for(; i <length - 1; ++i){
19                 if(Inorder.get(i).val > Inorder.get(i + 1).val){
20                     firstViolation = i;
21                     break;
22                 }                    
23             }
24             ++i;
25             for(; i < length - 1; ++i){
26                 if(Inorder.get(i).val > Inorder.get(i + 1).val){
27                     secondViolation = i + 1;
28                     break;
29                 }                    
30             }
31         }
32         if(secondViolation == 0)
33             secondViolation = firstViolation + 1;
34         int temp = Inorder.get(firstViolation).val;
35         Inorder.get(firstViolation).val = Inorder.get(secondViolation).val;
36         Inorder.get(secondViolation).val = temp;        
37     }
38     public ArrayList<TreeNode> InOrderTraversal(TreeNode root){
39         ArrayList<TreeNode> InOrder = new ArrayList<TreeNode>();
40         if(root != null){
41             if(root.left != null){
42                 ArrayList<TreeNode> leftChild = InOrderTraversal(root.left);
43                 InOrder.addAll(leftChild);
44             }
45             InOrder.add(root);
46             if(root.right != null){
47                 ArrayList<TreeNode> rightChild = InOrderTraversal(root.right);
48                 InOrder.addAll(rightChild);
49             }
50         }
51         return InOrder;
52     }
53 }

Solution2: using O(1) extra space:

 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 public class Solution {
11     public void recoverTree(TreeNode root) {
12         if(root != null){
13             inOrder(root);
14             int temp = firstViolation.val;
15             firstViolation.val = secondViolation.val;
16             secondViolation.val = temp;
17         }
18     }
19     
20     public void inOrder(TreeNode root){
21         if(root != null){
22             if(root.left != null)
23                 inOrder(root.left);
24             if(previous == null)
25                 previous = root;
26             else{
27                 if(previous.val > root.val){
28                     if(firstViolation == null)
29                         firstViolation = previous;
30                     secondViolation = root;                        
31                 }
32                 previous = root;
33             }            
34             if(root.right != null)
35                 inOrder(root.right);
36         }
37     }    
38     TreeNode previous = null;
39     TreeNode firstViolation = null;
40     TreeNode secondViolation = null;
41 }

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原文地址：https://www.cnblogs.com/averillzheng/p/3536156.html