leetcode 108. Convert Sorted Array to Binary Search Tree-数组转搜索树|递归|非递归

原题链接：108. Convert Sorted Array to Binary Search Tree
【思路-Java、Python】-递归实现

首先要明白二叉搜索树的性质——任意一个节点的值都不小于它的左子树上任一节点的值，任意一个节点的值不大于它的右子树上任一节点的值。高度平衡的二叉搜索数，要求平衡因子不大于1。

题目给定一个已经排好序的数组，要求构造一个高度平衡的二叉搜索树。那么，我们可以采用递归方式进行处理

1. 先得到数组元素的中间元素，将它的值作为根节点

2. 并且以它为中间中介，将数组划分为左右区间。根节点的左子树指向左区间，右节点指向右区间。

3. 再针对左右区间进行同样的操作。直到左指针和右指针重合。

public class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
return dfs(nums, 0, nums.length-1);
}
public TreeNode dfs(int[] nums, int left, int right) {
if(left > right) return null;
int mid = (left+right) / 2;
TreeNode root = new TreeNode(nums[mid]);
root.left = dfs(nums, left, mid-1);
root.right = dfs(nums, mid+1, right);
return root;
}
}

32 / 32 test
cases passed. Runtime: 1
ms  Your runtime beats 7.24% of javasubmissions.

class Solution(object):
def sortedArrayToBST(self, nums):
"""
:type nums: List[int]
:rtype: TreeNode
"""
def dfs(nums,left,right) :
if left <= right :
mid = (left+right) / 2
root = TreeNode(nums[mid])
root.left = dfs(nums,left,mid-1)
root.right = dfs(nums,mid+1,right)
return root
return dfs(nums,0,len(nums)-1)

32 / 32 test
cases passed. Runtime: 108
ms  Your runtime beats 23.81% of pythonsubmissions.

【思路2-Java】-非递归实现
可能有些读者想了解非递归方式是怎么实现的。这里我对递归实现做一个补充。

我们需要2个栈，一个栈用于存放区间的左、右边界指针——left、right，还需要一个栈存放当前子区间对应子树的根节点。相比递归实现，增加了很多变量的压栈、出栈操作，耗费时间也增加了不少：

public class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
if (nums == null || nums.length == 0) return null;
Stack<Integer> stack = new Stack<Integer>();
Stack<TreeNode> tree = new Stack<TreeNode>();
stack.add(nums.length - 1);
stack.add(0);
TreeNode root = new TreeNode(0);
tree.add(root);

while(!stack.isEmpty()) {
int left = stack.pop();
int right = stack.pop();
int mid = left + (right-left) / 2;
TreeNode node = tree.pop();
node.val = nums[mid];
int r = mid - 1, l = left;
if (l <= r) {
node.left = new TreeNode(0);
tree.add(node.left);
stack.push(r);
stack.push(l);
}
l = mid + 1;
r = right;
if (l <= r) {
node.right = new TreeNode(0);
tree.add(node.right);
stack.push(r);
stack.add(l);
}
}
return root;
}

32 / 32 test
cases passed. Runtime: 11
ms  Your runtime beats 0.95% of javasubmissions.