[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）。

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param A: Given an integer array with no duplicates.
     * @return: The root of max tree.
     */
    TreeNode* maxTree(vector<int> A) {
        // write your code here
        TreeNode *a, *b, *tmp;
        stack<TreeNode*> stk;
        int cur_max = INT_MIN;
        for (int i = 0; i <= A.size(); ++i) {
            if (i < A.size() && A[i] < cur_max) {
                tmp = new TreeNode(A[i]);
                stk.push(tmp);
            } else {
                b = NULL;
                while (!stk.empty() && (i == A.size() || stk.top()->val < A[i])) {
                    b = stk.top(); stk.pop();
                    if (!stk.empty()) a = stk.top()->right = b;
                }
                if (i < A.size()) {
                    tmp = new TreeNode(A[i]);
                    tmp->left = b;
                    stk.push(tmp);  
                } else {
                    stk.push(b);
                }
            }
        }
        return stk.top();
    }
};