再次完成98

Author: diguage

README.adoc

@@ -694,7 +694,7 @@
 |{source_base_url}/_0098_ValidateBinarySearchTree.java[Java]
 |{doc_base_url}/0098-validate-binary-search-tree.adoc[Note]
 |Medium
-|
+|树形DP套路或中序遍历
 
 //|99
 //|{leetcode_base_url}/recover-binary-search-tree/[Recover Binary Search Tree]

docs/0000-21-dynamic-programming.adoc

@@ -34,6 +34,10 @@ TIP: 写出信息结构是把所有的信息都装到一个对象中。如果只
 +
 . 第四步：设计递归函数，递归函数是处理以 X 为头节点的情况下的大难，包括设计递归的 base case，默认直接得到左树和右树的所有信息，以及把可能性做整合，并且要返回第三步的信息结构。
 
+=== 相关试题
+
+. xref:0098-validate-binary-search-tree.adoc[98. Validate Binary Search Tree]
+
 == 参考资料
 
 . 左程云《程序员代码面试指南》

docs/0098-validate-binary-search-tree.adoc

@@ -1,3 +1,4 @@
+[#0098-validate-binary-search-tree]
 = 98. Validate Binary Search Tree
 
 https://leetcode.com/problems/validate-binary-search-tree/[LeetCode - Validate Binary Search Tree]
@@ -54,3 +55,26 @@ Assume a BST is defined as follows:
 include::{sourcedir}/_0098_ValidateBinarySearchTree.java[]
 ----
 
+
+[{java_src_attr}]
+----
+include::{sourcedir}/_0098_ValidateBinarySearchTree_2.java[]
+----
+
+直接使用“树形DP套路”+剪枝技巧，速度直接击败 100%。
+
+这里有一点需要注意：最大值最小值用 `Long.MIN_VALUE` 和 `Long.MAX_VALUE`，这样可以防止单节点树 `Integer.MAX_VALUE` （最小值的单节点树应该也会有问题）造成的错误。
+
+另外，查看了官方题解后，发现可以使用树的中序排列来检查（二叉搜索树中序排列是升序），这样跟前几天在牛客网上做的那个“发现二叉搜索树中的两个错误节点”的思路就一致了。回头尝试一下。
+
+image::images/0098-01.png[]
+
+image::images/0098-02.png[]
+
+image::images/0098-03.png[]
+
+image::images/0098-04.png[]
+
+== 参考资料
+
+. https://leetcode.cn/problems/validate-binary-search-tree/solutions/230256/yan-zheng-er-cha-sou-suo-shu-by-leetcode-solution/[98. 验证二叉搜索树 - 官方题解^]

docs/images/0098-01.png

@@ -0,0 +1,64 @@
+package com.diguage.algorithm.leetcode;
+
+import com.diguage.algorithm.util.TreeNode;
+
+import java.util.Arrays;
+
+import static com.diguage.algorithm.util.TreeNodeUtils.buildTree;
+
+public class _0098_ValidateBinarySearchTree_2 {
+  public boolean isValidBST(TreeNode root) {
+    return dfs(root).valid;
+  }
+
+  private Record dfs(TreeNode node) {
+    if (node == null) {
+      return new Record(true, Long.MIN_VALUE, Long.MAX_VALUE);
+    }
+    Record left = dfs(node.left);
+    if (!left.valid) {
+      return new Record(false, Long.MIN_VALUE, Long.MAX_VALUE);
+    }
+    Record right = dfs(node.right);
+    if (!right.valid) {
+      return new Record(false, Long.MIN_VALUE, Long.MAX_VALUE);
+    }
+
+    long max = Math.max(left.max, Math.max(node.val, right.max));
+    long min = Math.min(left.min, Math.min(node.val, right.min));
+
+    return new Record(left.valid && right.valid
+      && left.max < node.val && node.val < right.min, max, min);
+  }
+
+  public static class Record {
+    public boolean valid;
+    public long max;
+    public long min;
+
+    public Record(boolean valid, long max, long min) {
+      this.valid = valid;
+      this.max = max;
+      this.min = min;
+    }
+  }
+
+  public static void main(String[] args) {
+    _0098_ValidateBinarySearchTree_2 solution = new _0098_ValidateBinarySearchTree_2();
+
+    boolean r5 = solution.isValidBST(buildTree(Arrays.asList(2147483647)));
+    System.out.println(r5);
+
+    boolean r4 = solution.isValidBST(buildTree(Arrays.asList(3, 1, 5, 0, 2, 4, 6, null, null, null, 3)));
+    System.out.println(r4);
+
+    boolean r3 = solution.isValidBST(buildTree(Arrays.asList(10, 5, 15, null, null, 6, 20)));
+    System.out.println(r3);
+
+    boolean r1 = solution.isValidBST(buildTree(Arrays.asList(2, 1, 3)));
+    System.out.println(r1);
+
+    boolean r2 = solution.isValidBST(buildTree(Arrays.asList(5, 1, 4, null, null, 3, 6)));
+    System.out.println(r2);
+  }
+}