代码随想录二刷day21

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前言

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一、力扣530. 二叉搜索树的最小绝对差

递归双指针法

/**
 * 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 {
    TreeNode pre = null;
    int diff = Integer.MAX_VALUE;
    public int getMinimumDifference(TreeNode root) {
        inOrder(root);
        return diff;
    }
    public void inOrder(TreeNode root){
        if(root == null){
            return ;
        }
        inOrder(root.left);
        if(pre == null){
            pre = root;
        }else{
            diff = Math.min(Math.abs(pre.val - root.val), diff);
            pre = root;
        }
        inOrder(root.right);
    }
}

迭代双指针法

/**
 * 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 getMinimumDifference(TreeNode root) {
        TreeNode pre = null;
        int diff = Integer.MAX_VALUE;
        Deque<TreeNode> deq = new LinkedList<>();
        TreeNode p = root;
        while(!deq.isEmpty() || p != null){
            if(p != null){
                deq.offerLast(p);
                p = p.left;
            }else{
                p = deq.pollLast();
                if(pre == null){
                    pre = p;
                }else{
                    diff = Math.min(diff, Math.abs(pre.val - p.val));
                    pre = p;
                }
                p = p.right;
            }
        }
        return diff;
    }
}

二、力扣501. 二叉搜索树中的众数

递归

/**
 * 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;
    int maxCount;
    int count;
    TreeNode pre;

    public int[] findMode(TreeNode root) {
        resList = new ArrayList<>();
        maxCount = 0;
        count = 0;
        pre = null;
        findMode1(root);
        int[] res = new int[resList.size()];
        for (int i = 0; i < resList.size(); i++) {
            res[i] = resList.get(i);
        }
        return res;
    }

    public void findMode1(TreeNode root) {
        if (root == null) {
            return;
        }
        findMode1(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;

        findMode1(root.right);
    }
}

三、力扣236. 二叉树的最近公共祖先

/**
 * 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 == p || root == q || root == null)return root;
        TreeNode leftChild = lowestCommonAncestor(root.left, p, q);
        TreeNode rightChild = lowestCommonAncestor(root.right, p, q);
        if(leftChild == null && rightChild != null){
            return rightChild;
        }else if(rightChild == null && leftChild != null){
            return leftChild;
        }else if(rightChild != null &&rightChild != null){
            return root;
        }else{
            return null;
        }
    }
}