leetcode 654. Maximum Binary Tree 递归构造最大二叉树 + 深度优先遍历DFS

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].

这个是典型的递归构造二叉树，按照定义构造即可

建议和leetcode 331. Verify Preorder Serialization of a Binary Tree 二叉树的前序序列验证 和 leetcode 297. Serialize and Deserialize Binary Tree 二叉树的序列化和反序列化 一起学习

下面是C++的做法：

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;


/*
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
*/

class Solution
{
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums)
    {
        if (nums.size() <= 0)
            return NULL;
        else
        {
            int maxNum = INT_MIN, index = -1;
            for (int i = 0; i < nums.size(); i++)
            {
                if (maxNum < nums[i])
                {
                    maxNum = nums[i];
                    index = i;
                }
            }
            vector<int> left(nums.begin(), nums.begin() + index);
            vector<int> right(nums.begin() + index + 1, nums.end());
            TreeNode* root = new TreeNode(maxNum);
            root->left = constructMaximumBinaryTree(left);
            root->right = constructMaximumBinaryTree(right);
            return root;
        }
    }
};