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Create Convert Sorted Array to Binary Search Tree

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	`1`	`+想法:用 Divide and Conquer 的技巧,把目前考慮的矩陣中間作為 root,其餘左右再分別構建子樹即可`
	`2`	`+終止條件是如果現在已經沒有矩陣需考慮就回傳 nullptr`
	`3`	`+`
	`4`	`+Time Complexity : O(n) for constructing n nodes`
	`5`	`+Space Complexity : O(n) for the binary tree`
	`6`	`+`
	`7`	`+/**`
	`8`	`+ * Definition for a binary tree node.`
	`9`	`+ * struct TreeNode {`
	`10`	`+ * int val;`
	`11`	`+ * TreeNode *left;`
	`12`	`+ * TreeNode *right;`
	`13`	`+ * TreeNode() : val(0), left(nullptr), right(nullptr) {}`
	`14`	`+ * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}`
	`15`	`+ * TreeNode(int x, TreeNode left, TreeNode right) : val(x), left(left), right(right) {}`
	`16`	`+ * };`
	`17`	`+ */`
	`18`	`+class Solution {`
	`19`	`+private:`
	`20`	`+ TreeNode* constructTree(vector<int>& nums , int startindex , int size) {`
	`21`	`+ if (size == 0)`
	`22`	`+ return nullptr ;`
	`23`	`+`
	`24`	`+ int middle = startindex + size / 2 ;`
	`25`	`+ return new TreeNode(nums[middle] ,`
	`26`	`+ constructTree(nums , startindex , middle - startindex) ,`
	`27`	`+ constructTree(nums , middle + 1 , startindex + size - middle - 1)) ;`
	`28`	`+ }`
	`29`	`+public:`
	`30`	`+ TreeNode* sortedArrayToBST(vector<int>& nums) {`
	`31`	`+ return constructTree(nums , 0 , nums.size()) ;`
	`32`	`+ }`
	`33`	`+};`

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