代码随想录刷题第19天

代码随想录刷题第19天

654. 最大二叉树

/**
 * 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 * transation(vector<int>& nums){
       
        int tmp = nums[0];
        int index = 0;
        for(int i = 1; i < nums.size(); i++){
            if(nums[i] > tmp){
                tmp = nums[i];
                index = i;
            }
        }
        TreeNode * root = new TreeNode(nums[index]);

        vector<int> leftboard(nums.begin(),nums.begin()+index);//左闭右开的初始化区间
        vector<int> rightboard(nums.begin()+index+1,nums.end());

        if(leftboard.size()!=0){
            root->left =transation(leftboard);
        }
        if(rightboard.size()!=0){
             root->right = transation(rightboard);
        }
       
        return root;
    }
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
       TreeNode *a = transation(nums);
       return a;
    }
};

617. 合并二叉树

/**
 * 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* traversal(TreeNode* root1, TreeNode* root2){
        if(root1 && !root2) return root1;
        else if(!root1 && root2) return root2;
        else if(!root2 && !root1) return nullptr;
        int value = root2->val + root1->val;
        TreeNode *node = new TreeNode(value);
        
       node->left= traversal(root1->left, root2->left);
       node->right= traversal(root1->right, root2->right);
        return node;

    }

    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
       TreeNode *result = traversal(root1,root2);
       return result;
    }
};

700. 二叉搜索树中的搜索

/**
 * 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 *result = nullptr;
    void traversal(TreeNode* root, int val){
        if(root == nullptr) return ;
        if(root->val == val) {
            result = root;
        }
        traversal(root->left, val);
        traversal(root->right, val);

    }
    TreeNode* searchBST(TreeNode* root, int val) {
        traversal(root, val);
        return  result;  
    }
};

98. 验证二叉搜索树

/**
 * 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> nums;
    void traversal(TreeNode *root){
        if(root == nullptr) return ;
        traversal(root->left);
        nums.push_back(root->val);
        traversal(root->right);
    }

    bool isValidBST(TreeNode* root) {
        bool result = true;
        traversal(root);
        for(int i = 1; i < nums.size(); i++){
            if(nums[i] <= nums[i-1]){
                result = false;
            }
        }
        return result;
    }
};