代码随想录算法训练营第十八天|LeetCode 530，501，236

目录

LeetCode  530.二叉搜索树的最小绝对差 

LeetCode 501.二叉搜索树中的众数 

LeetCode 236.二叉树的最近公共祖先

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LeetCode  530.二叉搜索树的最小绝对差 

文章讲解：代码随想录

视频讲解：二叉搜索树中，需要掌握如何双指针遍历！| LeetCode：530.二叉搜索树的最小绝对差_哔哩哔哩_bilibili

力扣题目：LeetCode  530.二叉搜索树的最小绝对差 

 

代码如下（Java）：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 {

    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);
    }
}

 代码如下（Java）：迭代法

/**
 * 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;
    Stack<TreeNode> stack;

    public int getMinimumDifference(TreeNode root) {

        if(root == null)    return 0;

        stack = new Stack<>();

        TreeNode cur = root;

        int result = Integer.MAX_VALUE;

        while(cur != null || !stack.isEmpty()){

            if(cur != null){
                stack.push(cur);
                cur = cur.left;
            }else{
                cur = stack.pop();
                if(pre != null){
                    result = Math.min(result, cur.val - pre.val);
                }

                pre = cur;
                cur = cur.right;
            }
        }
        return result;
    }

}

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LeetCode 501.二叉搜索树中的众数 

文章讲解：代码随想录

视频讲解：不仅双指针，还有代码技巧可以惊艳到你！ | LeetCode：501.二叉搜索树中的众数_哔哩哔哩_bilibili

力扣题目：LeetCode 501.二叉搜索树中的众数 

代码如下（Java）：中序遍历

/**
 * 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;

        findModel(root);

        int[] res = new int[resList.size()];

        for(int i = 0; i < resList.size(); i++){
            res[i] = resList.get(i);
        }

        return res;
    }

    public void findModel(TreeNode root){

        if(root == null)    return;

        findModel(root.left);

        int rootValue = root.val;

        if(pre == null || rootValue != pre.val){
            count = 1;
        }else{
            count++;
        }

        if(count > maxCount){
            resList.clear();
            resList.add(rootValue);
            maxCount = count;
        }else if(count == maxCount){
            resList.add(rootValue);
        }

        pre = root;

        findModel(root.right);
    }
}

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LeetCode 236.二叉树的最近公共祖先

 文章讲解：代码随想录

视频讲解：自底向上查找，有点难度！ | LeetCode：236. 二叉树的最近公共祖先_哔哩哔哩_bilibili

力扣题目：LeetCode 236.二叉树的最近公共祖先

 

 代码如下（Java）：后序遍历

/**
 * 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;

        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;
    }
}