[leetcode-654-Maximum Binary Tree]

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

1. The root is the maximum number in the array.
2. The left subtree is the maximum tree constructed from left part subarray divided by the maximum number.
3. 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:

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

思路：

用一个map来存储对应的数值和下标，正好利用map里面值是有序的特性，map里面最后一个值即为最大值。然后递归构造左子树和右子树。

TreeNode* maxtree(vector<int>& nums,int left,int right)
{
  if(left>right || nums.size()==0 || left<0 || right>=nums.size())return NULL;
  map<int,int>mp;
  for(int i=left;i<=right;i++)mp[nums[i]] =i;
  auto it = mp.rbegin();
  int maxnum = it->first;
  TreeNode* root = new TreeNode(maxnum);
  
   TreeNode* leftT = maxtree(nums,left,it->second -1);
   TreeNode* rightT = maxtree(nums,it->second +1,right);
   root->left = leftT,root->right =rightT;
  return root;
}
TreeNode* constructMaximumBinaryTree(vector<int>& nums)
{
   return maxtree(nums,0,nums.size()-1);
}

[leetcode-654-Maximum Binary Tree]

原文：http://www.cnblogs.com/hellowooorld/p/7294195.html