LeetCode Convert Sorted Array to Binary Search Tree

1.题目

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

2.解决方案1

 struct Node{     TreeNode* t;     int l;     int r;     Node(vector<int> &num, int l, int r){         this->t = new TreeNode(num[l+(r-l)/2]);         this->l = l;         this->r = r;     }     Node* getLeft(vector<int> &num){         int m = l + (r-l)/2;         if(l>=m) return NULL;         Node* left = new Node(num, l, m);         t->left = left->t;         return left;     }     Node* getRight(vector<int> &num){         int m = l + (r-l)/2;         if(m+1>=r) return NULL;         Node* right = new Node(num, m+1, r);         t->right = right->t;         return right;     } };class Solution {public:    TreeNode *sortedArrayToBST(vector<int> &num) {        if(num.size()==0) return NULL;        stack<Node*> s;        Node* root = new Node(num, 0, num.size());        s.push(root);        while(!s.empty()){            Node* n = s.top(); s.pop();            Node* left = n->getLeft(num);            if(left) s.push(left);            Node* right = n->getRight(num);            if(right) s.push(right);        }        return root->t;    }};

思路:把一个已经排好序的数组转换成一个平衡二叉搜索树。因为是平衡的二叉搜索树，即左右树的高度只相差1.所以对数组进行二分查找，这样找出的数才好建立树。非递归的方式要用到一个stack作为铺助，把所有的点都处理一次，树就建立好了。

3.解决方案2

class Solution {public:    TreeNode* build(vector<int> &num, int left, int right){if(left <= right){int mid = (left + right) / 2;TreeNode* rootNode = new TreeNode(num[mid]);rootNode->left = build(num,left,mid - 1);rootNode->right = build(num,mid + 1, right);return rootNode;}else{return NULL;}}    TreeNode *sortedArrayToBST(vector<int> &num) {        int len = num.size();TreeNode * root = NULL;if(len>0)root = build(num, 0, len-1);return root;    }};

思路:递归建立树是比较常用的方法，但速度非常慢。

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