LeetCode

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1
      / \
     2   3

Return 6.

public class Solution {

	int maxvalue = Integer.MIN_VALUE;
	
	public int getMaxSum(TreeNode root){
	   int maxleft=0;
	   int maxright=0;
       if(root.left!=null){
    	   maxleft = getMaxSum(root.left);
       }
       if(root.right!=null){
    	   maxright = getMaxSum(root.right);
       }
       int sum = root.val;
       if(maxleft>0){
    	   sum += maxleft;
       }
       if(maxright>0){
    	   sum +=maxright;
       }
       maxvalue = Math.max(maxvalue, sum);
       return Math.max(maxleft, maxright)>0?Math.max(maxleft, maxright)+root.val:root.val;
	}
	public int maxPathSum(TreeNode root) {
		
		int value = getMaxSum(root);
		
		value = Math.max(value, maxvalue);
		
		return value;
	}
	public static void main(String [] args){
		Solution solution = new Solution();
		TreeNode root = new TreeNode(1);
		root.left = new TreeNode(-2);
		root.right = new TreeNode(-3);
		
		root.left.left = new TreeNode(1);
		root.left.right = new TreeNode(3);
		root.left.left = new TreeNode(-1);
		
		root.right.left = new TreeNode(-2);
		//root.right.right = new TreeNode(3);
		//root.right.right.right = new TreeNode(-2);
		System.out.println(solution.maxPathSum(root));
	}
}