LeetCode_108.将有序数组转换为二叉搜索树

迷南。 2023-02-26 10:26 13阅读 0赞

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题解\_C

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     struct TreeNode *left;
     *     struct TreeNode *right;
     * };
     */
    
    //方法一：递归法
    //1,定位根节点
    //2,根节点左边作为左支递归处理
    //3,根节点右边作为右支递归处理
    
    struct TreeNode* sortedArrayToBST(int* nums, int numsSize){
        
        int     iRoot       = 0;
        struct TreeNode*    pCurNode    = NULL;
    
        //1,结束条件
        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,定位根节点
        iRoot = numsSize / 2;
    
        //4,递归处理左右支
        pCurNode->val = nums[iRoot];
        pCurNode->left = sortedArrayToBST(&nums[0], iRoot);
        pCurNode->right = sortedArrayToBST(&nums[iRoot + 1], numsSize - iRoot - 1);
        return pCurNode;
    }

题解\_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;
            }
    
            // 总是选择中间位置左边的数字作为根节点
            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;
        }
    }

题解\_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;
            } 
            // 以升序数组的中间元素作为根节点 root。
            int mid = lo + (hi - lo) / 2;
            TreeNode root = new TreeNode(nums[mid]);
            // 递归的构建 root 的左子树与右子树。
            root.left = dfs(nums, lo, mid - 1);
            root.right = dfs(nums, mid + 1, hi); 
            return root;
        }
    }

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