【小Y学算法】⚡️每日LeetCode打卡⚡️——29.将有序数组转换为二叉搜索树

public class Solution {
    static bool func(TreeNode x, TreeNode y) {
        if (x == null) {
            return y == null;
        }
        if (y == null || x.val != y.val) {
            return false;
        }
        return func(x.left, y.right) && func(x.right, y.left);
    }
    public bool IsSymmetric(TreeNode root) {
        return root == null ? true : func(root.left, root.right);
    }
}

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;
    }
}

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int p1 = 0, p2 = 0;
        int[] sorted = new int[m + n];
        int cur;
        while (p1 < m || p2 < n) {
            if (p1 == m) {
                cur = nums2[p2++];
            } else if (p2 == n) {
                cur = nums1[p1++];
            } else if (nums1[p1] < nums2[p2]) {
                cur = nums1[p1++];
            } else {
                cur = nums2[p2++];
            }
            sorted[p1 + p2 - 1] = cur;
        }
        for (int i = 0; i != m + n; ++i) {
            nums1[i] = sorted[i];
        }
    }
}