50 Maximum Binary Tree

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.

      6
    /   \
   3     5
    \    / 
     2  0   
      \
       1

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

找到当前数组最大值，左边的划为左子树数组，右边的划为右子树数组，最大值作为根。
递归重复上述操作。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        if(nums.length==0)
            return null;
        int max=maxIndex(nums);
        int[] left = new int[max];
        int[] right = new int[nums.length-max-1];
        // 左子树
        for(int i=0;i<left.length;++i){
            left[i]=nums[i];
        }
        // 右子树
        for(int i=0;i<right.length;++i){
            right[i]=nums[max+1+i];
        }
        TreeNode res = new TreeNode(nums[max]);
        res.left=constructMaximumBinaryTree(left);
        res.right=constructMaximumBinaryTree(right);
        return res;
    }
    
    int maxIndex(int[] arr){
        int res=0;
        for(int i=0;i<arr.length;++i){
            if(arr[i]>arr[res])
                res=i;
        }
        return res;
    }
}

public class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return construct(nums, 0, nums.length);
    }
    public TreeNode construct(int[] nums, int l, int r) {
        if (l == r)
            return null;
        int max_i = max(nums, l, r);
        TreeNode root = new TreeNode(nums[max_i]);
        root.left = construct(nums, l, max_i);
        root.right = construct(nums, max_i + 1, r);
        return root;
    }
    public int max(int[] nums, int l, int r) {
        int max_i = l;
        for (int i = l; i < r; i++) {
            if (nums[max_i] < nums[i])
                max_i = i;
        }
        return max_i;
    }
}