[LintCode] Max Tree

Max Tree

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

• The root is the maximum number in the array
• The left subtree and right subtree are the max trees of the subarray divided by the root number.

Construct the max tree by the given array.

 

Example

Given [2, 5, 6, 0, 3, 1], the max tree constructed by this array is:

    6

   / \

  5   3

 /   / \

2   0   1

Challenge

O(n) time and memory.

 

单调栈，每遍历到一个元素，将栈中小于该元素的结点合并，将合并后新的结点挂在当前结点的左侧。合并的时候，因为栈中的结点的值是递减的，所以直接将第二个结点挂下第一个结点的右侧。另外，这种树有个学名叫做笛卡树（Cartesian tree）。

 1 /**

 2  * Definition of TreeNode:

 3  * class TreeNode {

 4  * public:

 5  *     int val;

 6  *     TreeNode *left, *right;

 7  *     TreeNode(int val) {

 8  *         this->val = val;

 9  *         this->left = this->right = NULL;

10  *     }

11  * }

12  */

13 class Solution {

14 public:

15     /**

16      * @param A: Given an integer array with no duplicates.

17      * @return: The root of max tree.

18      */

19     TreeNode* maxTree(vector<int> A) {

20         // write your code here

21         TreeNode *a, *b, *tmp;

22         stack<TreeNode*> stk;

23         int cur_max = INT_MIN;

24         for (int i = 0; i <= A.size(); ++i) {

25             if (i < A.size() && A[i] < cur_max) {

26                 tmp = new TreeNode(A[i]);

27                 stk.push(tmp);

28             } else {

29                 b = NULL;

30                 while (!stk.empty() && (i == A.size() || stk.top()->val < A[i])) {

31                     b = stk.top(); stk.pop();

32                     if (!stk.empty()) a = stk.top()->right = b;

33                 }

34                 if (i < A.size()) {

35                     tmp = new TreeNode(A[i]);

36                     tmp->left = b;

37                     stk.push(tmp);  

38                 } else {

39                     stk.push(b);

40                 }

41             }

42         }

43         return stk.top();

44     }

45 };