LeetCode Day21 530&501&236

530. 二叉搜索树的最小绝对差二叉搜索树采用中序遍历，其实就是一个有序数组。对这棵搜索树进行排序以后求出最小差值即可。

class Solution {
    TreeNode pre;
    int result = Integer.MAX_VALUE;
    public int getMinimumDifference(TreeNode root) {
        if(root == null) return 0;
        traversal(root);
        return result;
    }

    public void traversal(TreeNode root){
        if(root == null) return;
        traversal(root.left);
        if(pre != null){
            result = Math.min(result, root.val-pre.val);
        }
        pre = root;
        traversal(root.right);
    }
}

501. 二叉搜索树中的众数

//暴力解题法
class Solution {
	public int[] findMode(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(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);
	}

}

236. 二叉树的最近公共祖先

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null || root == p || root == q){
            return root;
        }

        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if(left == null && right == null){
            return null;
        } else if(left == null && right != null){
            return right;
        } else if(left != null && right == null){
            return left;
        } else{
            return root;
        }
    }
}