題解_C

/*** Definition for a binary tree node.* struct TreeNode {* int val;* struct TreeNode *left;* struct TreeNode *right;* };*///方法一：遞歸法 //1,定位根節點 //2,根節點左邊作為左支遞歸處理 //3,根節點右邊作為右支遞歸處理struct TreeNode* sortedArrayToBST(int* nums, int numsSize){int iRoot = 0;struct TreeNode* pCurNode = NULL;//1,結束條件if((NULL == nums) || (0 == numsSize)) return NULL;//2,初始化pCurNode = (struct TreeNode*)malloc(sizeof(struct TreeNode)); // memset(pCurNode, 0x00, sizeof(struct TreeNode));//3,定位根節點iRoot = numsSize / 2;//4,遞歸處理左右支pCurNode->val = nums[iRoot];pCurNode->left = sortedArrayToBST(&nums[0], iRoot);pCurNode->right = sortedArrayToBST(&nums[iRoot + 1], numsSize - iRoot - 1);return pCurNode; }

題解_Java_1

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode(int x) { val = x; }* }*/ class Solution {public TreeNode sortedArrayToBST(int[] nums) {return helper(nums, 0, nums.length - 1);}public TreeNode helper(int[] nums, int left, int right) {if (left > right) {return null;}// 總是選擇中間位置左邊的數字作為根節點int mid = (left + right) / 2;TreeNode root = new TreeNode(nums[mid]);root.left = helper(nums, left, mid - 1);root.right = helper(nums, mid + 1, right);return root;} }

題解_Java_2

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode(int x) { val = x; }* }*/ class Solution {public TreeNode sortedArrayToBST(int[] nums) {return dfs(nums, 0, nums.length - 1);}private TreeNode dfs(int[] nums, int lo, int hi) {if (lo > hi) {return null;} // 以升序數組的中間元素作為根節點 root。int mid = lo + (hi - lo) / 2;TreeNode root = new TreeNode(nums[mid]);// 遞歸的構建 root 的左子樹與右子樹。root.left = dfs(nums, lo, mid - 1);root.right = dfs(nums, mid + 1, hi); return root;} } 與50位技術專家面對面20年技術見證，附贈技術全景圖

總結

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