day21 - 二叉树part07

530. 二叉搜索树的最小绝对差

详解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* pre = NULL;
    int result = INT_MAX;
    //中序遍历是递增的
    int getMinimumDifference(TreeNode* root) {
        if(root == NULL) return -1;
        getMinimumDifference(root->left);
        if(pre != NULL){
            result = min(result, abs(pre->val - root->val));
        }
        pre = root;
        getMinimumDifference(root->right);
        return result;
    }
};

 

501. 二叉搜索树中的众数

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> result;
    TreeNode* pre = NULL;
    int count=0, max_count=0;
    //中序遍历递增
    vector<int> findMode(TreeNode* root) {
        if(!root) return {};
        findMode(root->left);

        if(pre == NULL){
            count = 1;
        }else if(pre->val == root->val){//连续相同
            count++;
        }else{
            count = 1;
        }

        if (count == max_count) { // 如果和最大值相同，放进result中
            result.push_back(root->val);
        }

        pre = root;

        if (count > max_count) { // 如果计数大于最大值频率
            max_count = count;   // 更新最大频率
            result.clear();     // 很关键的一步，不要忘记清空result，之前result里的元素都失效了
            result.push_back(root->val);
        }

        findMode(root->right);

        return result;

    }
};

 

236. 二叉树的最近公共祖先

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == p || root == q || root == NULL) return root;
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);

        if(left && right) return root;

        return left? left:right;  //有1个是NULL
    }
};