package tree;

/**

* Created by gouthamvidyapradhan on 19/08/2017.

* Given an integer array with no duplicates. A maximum tree building on this array is defined as follow:

The root is the maximum number in the array.

The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.

The right subtree is the maximum tree constructed from right part subarray divided by the maximum number.

Construct the maximum tree by the given array and output the root node of this tree.

Example 1:

Input: [3,2,1,6,0,5]

Output: return the tree root node representing the following tree:

6

/ \

3 5

\ /

2 0

\

1

Note:

The size of the given array will be in the range [1,1000].

*/

public class MaximumBinaryTree {

public static class TreeNode {

int val;

TreeNode left;

TreeNode right;

TreeNode(int x) { val = x; }

}

private int[][] max;

/**

* Main method

* @param args

* @throws Exception

*/

public static void main(String[] args) throws Exception{

int[] nums = {3,2,1,6,0,5};

TreeNode root = new MaximumBinaryTree().constructMaximumBinaryTree(nums);

System.out.println(root.val); //print root

}

public TreeNode constructMaximumBinaryTree(int[] nums) {

max = new int[nums.length][nums.length];

//pre-fill with initial values

for(int i = 0; i < nums.length; i ++){

max[i][i] = i;

}

//pre-calculate max for range index

for(int i = 0; i < nums.length; i ++){

for(int j = i + 1; j < nums.length; j++){

max[i][j] = nums[max[i][j - 1]] > nums[j] ? max[i][j - 1] : j;

}

}

return build(0, nums.length - 1, nums);

}

private TreeNode build(int s, int e, int[] nums){

if(s <= e){

int val = nums[max[s][e]];

TreeNode n = new TreeNode(val);

n.left = build(s, max[s][e] - 1, nums);

n.right = build(max[s][e] + 1, e, nums);

return n;

}

return null;

}

}