Maximum Binary Tree

此题和利用前序加中序序列构建二叉树那道题类似，都是利用三个坐标去限定范围，再利用递归逐渐缩小范围最后完成构建，值得注意的是base case的判断条件，只要左右有一边的范围没有非法，我们就还需要继续向下构建子树，除非两边都非法了，这样才返回空节点，时间复杂度为O(n^2)

class Solution(object):
    def constructMaximumBinaryTree(self, nums):
        """
        :type nums: List[int]
        :rtype: TreeNode
        """
        return self.buildTree(nums, 0, nums.index(max(nums)), len(nums) - 1)

    def buildTree(self, nums, start, rootIndex, end):
        if start > rootIndex and end < rootIndex:
            return None
        left, right = None, None
        root = TreeNode(nums[rootIndex])
        if start < rootIndex:
            root.left = self.buildTree(nums, start, nums.index(max(nums[start : rootIndex])), rootIndex - 1)
        if end > rootIndex:
            root.right = self.buildTree(nums, rootIndex + 1, nums.index(max(nums[rootIndex + 1 : end + 1])), end)
        return root