Update 0501.二叉搜索树中的众数.md

java的暴力解法（map记录频率）
nowsoar · Jul 1, 2021 · 383c9cb · 383c9cb
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diff --git a/problems/0501.二叉搜索树中的众数.md b/problems/0501.二叉搜索树中的众数.md
@@ -345,6 +345,40 @@ public:
 
 Java：
 
+暴力法
+```java
+class Solution {
+	public int[] findMode(FindModeInBinarySearchTree.TreeNode root) {
+		Map<Integer, Integer> map = new HashMap<>();
+		List<Integer> list = new ArrayList<>();
+		if (root == null) return list.stream().mapToInt(Integer::intValue).toArray();
+		// 获得频率 Map
+		searchBST(root, map);
+		List<Map.Entry<Integer, Integer>> mapList = map.entrySet().stream()
+				.sorted((c1, c2) -> c2.getValue().compareTo(c1.getValue()))
+				.collect(Collectors.toList());
+		list.add(mapList.get(0).getKey());
+		// 把频率最高的加入 list
+		for (int i = 1; i < mapList.size(); i++) {
+			if (mapList.get(i).getValue() == mapList.get(i - 1).getValue()) {
+				list.add(mapList.get(i).getKey());
+			} else {
+				break;
+			}
+		}
+		return list.stream().mapToInt(Integer::intValue).toArray();
+	}
+
+	void searchBST(FindModeInBinarySearchTree.TreeNode curr, Map<Integer, Integer> map) {
+		if (curr == null) return;
+		map.put(curr.val, map.getOrDefault(curr.val, 0) + 1);
+		searchBST(curr.left, map);
+		searchBST(curr.right, map);
+	}
+
+}
+```
+
 ```Java
 class Solution {
     ArrayList<Integer> resList;