Binary Tree Longest Consequence

 1 /**
 2  * Definition for a binary tree node.
 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     private int result = 1;
12     public int longestConsecutive(TreeNode root) {
13         if (root == null) {
14             return 0;
15         }
16         getLongest(root, Integer.MIN_VALUE, 0);
17         return result;
18     }
19     
20     private void getLongest(TreeNode root, int target, int count) {
21         if (root == null) {
22             result = Math.max(result, count);
23             return;
24         }
25         
26         if (root.val == target + 1) {
27             count++;
28             result = Math.max(result, count);
29         } else {
30             count = 1;
31         }
32         getLongest(root.left, root.val, count);
33         getLongest(root.right, root.val, count);
34     }
35 }

Maximum value get should be count increase. Otherwise, it already breaks the consequences.

转载于:https://wwwblogs/shuashuashua/p/5645615.html

Binary Tree Longest Consequence

 1 /**
 2  * Definition for a binary tree node.
 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     private int result = 1;
12     public int longestConsecutive(TreeNode root) {
13         if (root == null) {
14             return 0;
15         }
16         getLongest(root, Integer.MIN_VALUE, 0);
17         return result;
18     }
19     
20     private void getLongest(TreeNode root, int target, int count) {
21         if (root == null) {
22             result = Math.max(result, count);
23             return;
24         }
25         
26         if (root.val == target + 1) {
27             count++;
28             result = Math.max(result, count);
29         } else {
30             count = 1;
31         }
32         getLongest(root.left, root.val, count);
33         getLongest(root.right, root.val, count);
34     }
35 }

Maximum value get should be count increase. Otherwise, it already breaks the consequences.

转载于:https://wwwblogs/shuashuashua/p/5645615.html