[LeetCode-Algorithms-654] "Maximum Binary Tree" (2017.9.8)

题目链接：Maximum Binary Tree

题目描述：

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:



Note:

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

<1>思路：循环数组每一个元素，首先申请当前数组元素的节点，然后和已进入树中的最后一个节点（后面称为记录节点）比较大小，如果小于树中记录的最后一个节点，就直接跟在当前树记录节点的右子树（保证每个子树小的在最下面）；否则，把记录节点设置为当前数字节点的左子节点，然后抛掉它并将父节点设置为记录节点，再和它的父节点比较大小，直到把当前这个数节点插入到正确位置。保证最大数二元树每个子树都符合要求，即最大的在最上面。这是一个时间复杂度为O(n)的做法。

<2>代码：

class Solution {
public:
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
vector<TreeNode*> con;
for (int i = 0; i < nums.size(); i++) {
TreeNode* curr = new TreeNode(nums[i]);

while (!con.empty() && con.back()->val < nums[i]) {
curr->left = con.back();
con.pop_back();
}

if (!con.empty())
con.back()->right = curr;
con.push_back(curr);
}
return con.front();
}
};

<3>提交结果：