代码随想录day21

一、二叉搜索树的最小绝对差（LeetCode530）

遇到在二叉搜索树上求什么最值，求差值之类的，都要思考一下二叉搜索树可是有序的，要利用好这一特点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int res = Integer.MAX_VALUE;
    public int getMinimumDifference(TreeNode root) {
        traversal(root);
        return res; 
    }
    TreeNode pre = null;// 记录上一个遍历的结点
    void traversal(TreeNode cur){
        if(cur == null) return;
        //中序遍历
        traversal(cur.left);
        if(pre != null){
            res = Math.min(res,cur.val - pre.val);
        }
        pre = cur;
        traversal(cur.right);
    }
}

二、二叉搜索树中的众数（LeetCode501）

中序遍历-不使用额外空间，利用二叉搜索树特性

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    ArrayList<Integer> resList = new ArrayList<>();
    int maxCount = 0;
    int count = 0;
    TreeNode pre = null;

    public int[] findMode(TreeNode root) {
        travesal(root);
        int[] res = new int[resList.size()];
        for (int i = 0; i < resList.size(); i++) {
            res[i] = resList.get(i);
        }
        return res;
    }

    public void travesal(TreeNode root) {
        if (root == null) {
            return;
        }
        travesal(root.left);
        int rootValue = root.val;
        // 计数
        if (pre == null || rootValue != pre.val) {
            count = 1;
        } else {
            count++;
        }
        // 更新结果以及maxCount
        if (count > maxCount) {
            resList.clear();
            resList.add(rootValue);
            maxCount = count;
        } else if (count == maxCount) {
            resList.add(rootValue);
        }
        pre = root;
        travesal(root.right);
    }
}

暴力求解

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
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);
	}
}

三、二叉树的最近公共祖先（LeetCode236）

递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) { // 递归结束条件
            return root;
        }
        // 后序遍历
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if(left == null && right == null) { // 若未找到节点 p 或 q
            return null;
        }else if(left == null && right != null) { // 若找到一个节点
            return right;
        }else if(left != null && right == null) { // 若找到一个节点
            return left;
        }else { // 若找到两个节点
            return root;
        }
    }
}