代码随想录训练营day19||654.最大二叉树 617.合并二叉树 700.二叉搜索树中的搜索 98.验证二叉搜索树-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_74108081/article/details/136143501

凡是构造二叉树的用前序遍历

/**
 * 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(vector<int>& nums,int left,int right)
    {
        if(left>=right)
        return NULL;
        int maxvalue=0;
        int index=left;
        for(int i=left;i<right;i++)
        {
            if(nums[i]>maxvalue)
            {
                maxvalue=nums[i];
               index=i;
            }
        }
        TreeNode* root=new TreeNode(nums[index]);
        root->left=traversal(nums,left,index);
        root->right=traversal(nums,1+index,right);
        return root;
    }
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
    return traversal(nums,0,nums.size());
    }
};

617. 合并二叉树 - 力扣（LeetCode）

一起操作两个二叉树？有点懵！| LeetCode：617.合并二叉树_哔哩哔哩_bilibili

/**
 * 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* mergeTrees(TreeNode* root1, TreeNode* root2) {    
        if(root1==NULL)
        return root2;
        if(root2==NULL)
        return root1;
        root1->val+=root2->val;
        root1->left=mergeTrees(root1->left,root2->left);
        root1->right=mergeTrees(root1->right,root2->right);
        return root1;
    }
};

700. 二叉搜索树中的搜索 - 力扣（LeetCode）

/**
 * 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* searchBST(TreeNode* root, int val) {
        if(root==NULL||root->val==val)
        return root;
        TreeNode* result=NULL;
        if(val<root->val)
        result=searchBST(root->left,val);
        if(val>root->val)
        result=searchBST(root->right,val);
        return result;
    }
};

98. 验证二叉搜索树 - 力扣（LeetCode）

/**
 * 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;
    bool isValidBST(TreeNode* root) {
        if(root==NULL)
        return true;
        bool left=isValidBST(root->left);
        if(pre!=NULL&&pre->val>=root->val)
        return false;
        pre=root;
        bool right=isValidBST(root->right);
        return left&&right;
    }
};

记住 左子树所有节点小于中间节点，右子树所有节点大于中间节点！！！

还有最小值该如何定义，样例中的最小值可能是int型的最小值，怎么去比较大小，所以我们采取引入pre节点，这样就不用定义最小值了