01.06.字符串压缩

✒解题思路：

  1. 代码通俗…
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class Solution {
	public String compressString(String S) {
		if (S == null || S.length() <= 2) {
			return S;
		}
		StringBuilder stringBuilder = new StringBuilder();
		int cnt = 1;
		stringBuilder.append(S.charAt(0));
		for (int i = 1; i < S.length(); i++) {
			if (S.charAt(i) == S.charAt(i - 1)) {
				cnt++;
			} else {
				stringBuilder.append(cnt).append(S.charAt(i));
				cnt = 1;
			}
		}
        stringBuilder.append(cnt);
		return stringBuilder.toString().length() > S.length() ? S : stringBuilder.toString();
	}
}

98.验证二叉搜索树

✒解题思路：

  1. 对于二叉搜索树，左节点必须小于根，右节点必须大于根，并且每个子树也是如此。
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
	public boolean isValidBST(TreeNode root) {
		return helper(root, null, null);
	}

	private boolean helper(TreeNode root, Integer lower, Integer upper) {
		if (root == null) {
			return true;
		}
		int val = root.val;
		if ((lower != null && val <= lower) || (upper != null && val >= upper)) {
			return false;
		}
		return helper(root.right, val, upper) && helper(root.left, lower, val);
	}
}