# 1130.Minimum-Cost-Tree-From-Leaf-Values

## 1130. Minimum Cost Tree From Leaf Values

## 题目地址

<https://leetcode.com/problems/minimum-cost-tree-from-leaf-values/>

## 题目描述

```
Given an array arr of positive integers, consider all binary trees such that:

Each node has either 0 or 2 children;
The values of arr correspond to the values of each leaf in an in-order traversal of the tree.  (Recall that a node is a leaf if and only if it has 0 children.)
The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree respectively.
Among all possible binary trees considered, return the smallest possible sum of the values of each non-leaf node.  It is guaranteed this sum fits into a 32-bit integer.

Example 1:
Input: arr = [6,2,4]
Output: 32
Explanation:
There are two possible trees.  The first has non-leaf node sum 36, and the second has non-leaf node sum 32.

    24            24
   /  \          /  \
  12   4        6    8
 /  \               / \
6    2             2   4


Constraints:
2 <= arr.length <= 40
1 <= arr[i] <= 15
It is guaranteed that the answer fits into a 32-bit signed integer (ie. it is less than 2^31).
```

## 代码

### Approach #1 单调递减栈

考虑使用一个单调递减栈来存储数组元素，如果当前元素小于等于栈顶元素直接入栈；否则，说明当前栈顶元素是一个局部最小值，我们用这个局部最小值的两端较小与栈顶元素相乘，同时将栈顶元素出栈，重复上述操作，直至栈顶元素不是局部最小值，再将当前元素入栈。处理完整个数组后，如果栈中还有多的元素，我们从栈顶依次往下乘加入答案即可

```java
class Solution {
  public int mctFromLeafValues(int[] arr) {
        int res = 0;
    Stack<Integer> stack = new Stack();
    stack.push(Integer.MAX_VALUE);
    for (int a : arr) {
      while (stack.peek() <= a) {
        int mid = stack.pop();
        res += mid * Math.min(stack.peek(), a);
      }
      stack,push(a);
    }

    while (stack.size() > 2) {
      res += stack.pop() * stack.peek();
    }

    return res;
  }
}
```

### Approach #2 Dynamic Programming

`dp[i][j]` = answer of build a tree from `a[i] ~ a[j]` `dp[i][j]` = `min{dp[i][k] + dp[k+1][j] + max(a[i~k]) * max(a[k+1~j])} i <= k < j`

```java
class Solution {
  public int mctFromLeafValues(int[] arr) {
    int n = arr.size();
    int[][] dp = new int[n][n];
    int[][] m = new int[n][n];
    for (int i = 0; i < n; i++) {
      m[i][i] = arr[i];
      for (int j = i + 1; j < n; j++) {
        m[i][j] = Math.max(m[i][j - 1], arr[j]);
      }
    }
    for (int l = 2; l <= n; i++) {
      for (int i = 0; i + l <= n; i++) {
        int j = i + l - 1;
        dp[i][j] = Integer.MAX_VALUE;
        for (int k = i; k < j; k++) {
          dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k + 1][j] + m[i][k] * m[k + 1][j]); 
        }
      }
    }
    return dp[0][n - 1];
  }
}
```


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