思路
- 题目要求构造平衡二叉树,也就是左右子树的节点数尽可能接近
- 这就转化成了递归建立二叉树的问题
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* sortedArrayToBST(vector<int>& nums) {
return helper(nums,0,nums.size());//一般设计成左闭右开
//这样处理起来比较方便,如果左闭右闭,两个相邻区间的交接点会有重复,处理起来就比较繁琐
}
TreeNode* helper(vector<int>& nums,int left,int right){
if(left == right) return NULL;
int mid= (left + right)>>1;
TreeNode* node = new TreeNode(nums[mid]);
node->left = helper(nums,left,mid);
node->right = helper(nums,mid+1,right);
return node;
}
};
这样写比较容易理解
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* sortedArrayToBST(vector<int>& nums) {
if(nums.size() == 0) return NULL;
return dfs(0,nums.size()-1,nums);
}
TreeNode* dfs(int left, int right,vector<int>& nums){
if(left <= right){
int mid = (left +right) >> 1; //这里写成left + (right-left)>> 1
TreeNode* node = new TreeNode(nums[mid]);
node->left = dfs(left,mid-1,nums);
node->right = dfs(mid+1,right,nums);
return node;
}
return NULL;
}
};