补-代码随想录第21天|● 530.二叉搜索树的最小绝对差 ● 501.二叉搜索树中的众数 ● 236. 二叉树的最近公共祖先

● 530.二叉搜索树的最小绝对差

思路一：递归-中序遍历

代码：定义双指针做插值

public class TreeNode{
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(){};
    TreeNode(int val,TreeNode left,TreeNode right){
        this.val=val;
        this.left=left;
        this.right=right;
    }
    TreeNode(int val){
        this.val=val;
    }
}
class Solution {
    int result=Integer.MAX_VALUE;
    TreeNode pre;//默认为null
    public int getMinimumDifference(TreeNode root) {
        if(root==null)return 0;
        traversal(root);
        return result;
    }
    public void traversal(TreeNode root){
        //中序遍历 左中右
        //终止条件
        if(root==null)return;
        traversal(root.left);//左
        //中
        if(pre!=null){
            // result=Math.min(Math.abs(root.val-pre.val),result);//不用绝对值，前-后一定递增
            result=Math.min(root.val-pre.val,result);
        }
        pre=root;
        traversal(root.right);

    }
}

思路二：统一迭代法-速度较慢

/**
 * 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 result=Integer.MAX_VALUE;
    TreeNode pre;//默认为null
    Stack<TreeNode> stack = new Stack<>();
    public int getMinimumDifference(TreeNode root) {
        if(root!=null){
            stack.push(root);
        }
        while(!stack.isEmpty()){
            TreeNode cur=stack.peek();
            if(cur!=null){
                stack.pop();
                if(cur.right != null)//右
                    stack.add(cur.right);
                stack.add(cur);//中
                stack.add(null);
                if(cur.left != null)//左
                    stack.add(cur.left);
            }else{
                stack.pop();
                cur=stack.pop();
                if(pre!=null)
                    result=Math.min(cur.val-pre.val,result);
                pre=cur;
            }
        }
        return result;
    }
}

● 501.二叉搜索树中的众数

思路：

代码：

class Solution {
    int count;
    ArrayList<Integer> resList;
    int maxcount;
    TreeNode pre;
    public int[] findMode(TreeNode root) {
        count=1;
        maxcount=Integer.MIN_VALUE;
        resList=new ArrayList<>();
        search(root);
        int n=resList.size();
        int[] res=new int[n];
        for(int i=0;i<n;i++){
            res[i]=resList.get(i);
        }
        return res;
    }
    public void search(TreeNode root) {
        if(root==null)return;
        search(root.left);
        if(pre!=null){
            if(pre.val==root.val){
                count++;
            }else{
                count=1;
            }
        }
        pre=root;
        if(count==maxcount){
            resList.add(root.val);
        }
        if(count>maxcount){
            maxcount=count;
            resList.clear();
            resList.add(root.val);
        }
        search(root.right);
    }

● 236. 二叉树的最近公共祖先

思路

情况1

情况2

递归三部曲

如图：

代码：

/**
 * 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 )return root;
        // 左右中
        TreeNode left=lowestCommonAncestor(root.left,p,q);
        TreeNode right=lowestCommonAncestor(root.right,p,q);
        //中
        if(left!=null&&right!=null){
            return root;
        }
        else if(left==null&&right!=null){
            return right;
        }else if(left!=null&&right==null){
            return left;
        }else{
            return null;
        }
    }
}