public static void main(String args[]){

TreeNode treeNode = new TreeNode(1);

treeNode.right = new TreeNode(2);

treeNode.right.right = new TreeNode(3);

treeNode.right.right.right = new TreeNode(4);

treeNode.right.right.right.right = new TreeNode(5);

System.out.printf("Max path sum: %d",

maxPathSum(treeNode)

);

int[] a = new int[3];

a[0] = 1;

a[2] = 3;

TEST(a);

System.out.printf("a: %d", a[0]);

//a = {0, 1, 2, 3};

ArrayList<Integer> a1 = new ArrayList<Integer>();

ArrayList<Integer> b1 = a1;

a1.add(1);

b1.add(2);

System.out.printf("size of a: %d", a1.size());

}

public static void TEST(int[] a){

a[0] = 4;

}

public static int maxPathSum(TreeNode root) {

// set the result to min value.

int result = Integer.MIN_VALUE;

if (root == null){

return result;

}

// divide the problem to be two parts. If it is smaller than 0, discard it.

int left_signal = Math.max(signal_Sum(root.left), 0);

int right_signal = Math.max(signal_Sum(root.right), 0);

System.out.printf("LeftSignal:%d right_signal:%d\n", left_signal, right_signal);

// get the max of the cross.

int cross_max = root.val + left_signal + right_signal;

int left_max = maxPathSum(root.left);

int right_max = maxPathSum(root.right);

System.out.printf("cross_max:%d left_max:%d right_max:%d\n", cross_max, left_max, right_max);

result = Math.max(cross_max, Math.max(left_max, right_max));

return result;

}

// return the path which include the root and one of the branch.

private static int signal_Sum(TreeNode root){

if (root == null){

return Integer.MIN_VALUE;

}

int result = Math.max(Math.max(signal_Sum(root.left), 0) + root.val, Math.max(signal_Sum(root.right), 0) + root.val);

System.out.printf("SIGNAL_SUM result:%d\n", result);

// if branch < 0, we can just return the root node value.

return result;

}