654.Maximum Binary Tree

654.Maximum Binary Tree

难度:Medium

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].

可以采用递归的方法，先序遍历。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
vector<int> num;
TreeNode* getMaxInd(int start, int end)
{
    if(start == end) return new TreeNode(num[start]);
    int m=INT_MIN;
    int res=start;
    for(int i=start;i<=end;i++)
    {
        if(num[i]>m) 
        {
            res=i;
            m=num[i];
        }
    }
    TreeNode* root= new TreeNode(num[res]);
    if(res-start>0)   root->left = getMaxInd(start, res-1);
    if(end - res>0) root->right = getMaxInd(res+1,end);
    return root;
}
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        num=nums;
        return getMaxInd(0,num.size()-1);


    }
};

执行用时 :136 ms, 在所有 C++ 提交中击败了60.70%的用户 内存消耗 :29.2 MB, 在所有 C++ 提交中击败了67.54%的用户