You are given an integer array nums with no duplicates. A maximum binary tree can be built recursively from nums using the following algorithm:
- Create a root node whose value is the maximum value in
nums. - Recursively build the left subtree on the subarray prefix to the left of the maximum value.
- Recursively build the right subtree on the subarray suffix to the right of the maximum value.
Return the maximum binary tree built from nums.
Example 1:
Input: nums = [3,2,1,6,0,5]
Output: [6,3,5,null,2,0,null,null,1]
Explanation: The recursive calls are as follow:
- The largest value in [3,2,1,6,0,5] is 6. Left prefix is [3,2,1] and right suffix is [0,5].
- The largest value in [3,2,1] is 3. Left prefix is [] and right suffix is [2,1].
- Empty array, so no child.
- The largest value in [2,1] is 2. Left prefix is [] and right suffix is [1].
- Empty array, so no child.
- Only one element, so child is a node with value 1.
- The largest value in [0,5] is 5. Left prefix is [0] and right suffix is [].
- Only one element, so child is a node with value 0.
- Empty array, so no child.
Example 2:
Input: nums = [3,2,1] Output: [3,null,2,null,1]
Constraints:
1 <= nums.length <= 10000 <= nums[i] <= 1000- All integers in
numsare unique.
给定一个不重复的整数数组 nums 。 最大二叉树 可以用下面的算法从 nums 递归地构建:
- 创建一个根节点，其值为
nums中的最大值。 - 递归地在最大值 左边 的 子数组前缀上 构建左子树。
- 递归地在最大值 右边 的 子数组后缀上 构建右子树。
返回 nums 构建的 最大二叉树 。
示例 1：
输入：nums = [3,2,1,6,0,5]
输出：[6,3,5,null,2,0,null,null,1]
解释：递归调用如下所示：
- [3,2,1,6,0,5] 中的最大值是 6 ，左边部分是 [3,2,1] ，右边部分是 [0,5] 。
- [3,2,1] 中的最大值是 3 ，左边部分是 [] ，右边部分是 [2,1] 。
- 空数组，无子节点。
- [2,1] 中的最大值是 2 ，左边部分是 [] ，右边部分是 [1] 。
- 空数组，无子节点。
- 只有一个元素，所以子节点是一个值为 1 的节点。
- [0,5] 中的最大值是 5 ，左边部分是 [0] ，右边部分是 [] 。
- 只有一个元素，所以子节点是一个值为 0 的节点。
- 空数组，无子节点。
示例 2：
输入：nums = [3,2,1] 输出：[3,null,2,null,1]
提示：
1 <= nums.length <= 10000 <= nums[i] <= 1000nums中的所有整数 互不相同
| Language | Runtime | Memory | Submission Time |
|---|---|---|---|
| javascript | 156 ms | 14.8 MB | 2018/12/23 17:52 |
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {number[]} nums
* @return {TreeNode}
*/
function divided(nums, root, leftIndex, rightIndex) {
if(leftIndex === rightIndex) {
root.val = nums[leftIndex];
return;
}
var maxIndex = -1, max = -1;
for (var i = leftIndex; i <= rightIndex; i++) {
if (nums[i] > max) {
max = nums[i];
maxIndex = i;
}
}
root.val = nums[maxIndex];
if (maxIndex-1 < leftIndex) {
root.right = new TreeNode(-100);
divided(nums, root.right, maxIndex+1, rightIndex);
} else if(maxIndex+1 > rightIndex) {
root.left = new TreeNode(-100);
divided(nums, root.left, leftIndex, maxIndex-1);
} else {
root.left = new TreeNode(-100);
root.right = new TreeNode(-100);
divided(nums, root.left, leftIndex, maxIndex-1);
divided(nums, root.right, maxIndex+1, rightIndex);
}
}
var constructMaximumBinaryTree = function(nums) {
var root = new TreeNode(-1);
divided(nums, root, 0, nums.length-1);
return root;
};