Leetcode 654-最大二叉树

题目描述

https://leetcode.cn/problems/maximum-binary-tree/description/

解题思路

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        TreeNode* root = new TreeNode(nums[0]);
        if (nums.size() == 1)return root;
        int maxIndex = 0;
        for (int i = 0; i < nums.size(); i++) {
            if (nums[i] > root->val) {
                root->val = nums[i];
                maxIndex = i;
            }
        }
        if (maxIndex > 0) {
            vector<int> left(nums.begin(), nums.begin() + maxIndex);
            root->left = constructMaximumBinaryTree(left);
        }
        if (maxIndex < nums.size() - 1) {
            vector<int> right(nums.begin() + maxIndex + 1, nums.end());
            root->right = constructMaximumBinaryTree(right);
        }
        return root;
    }
};

Leetcode 617-合并二叉树

题目描述

https://leetcode.cn/problems/merge-two-binary-trees/description/

解题思路

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        if (root1 == nullptr)return root2;
        if (root2 == nullptr) return root1;
        TreeNode* root = new TreeNode(0);//定义一个新的节点接收结果
        root->val = root1->val + root2->val;
        root->left = mergeTrees(root1->left, root2->left);
        root->right = mergeTrees(root1->right, root2->right);
        return root;
    }
};

Leetcode 700-二叉搜索树中的搜索

题目描述

https://leetcode.cn/problems/search-in-a-binary-search-tree/description/

解题思路

二叉搜索树：二叉搜索树自带顺序，根节点的左子树的节点要比根节点的值都小，右子树的节点的值比根节点的都大。

递归法：

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if (root == nullptr || root->val == val) return root;
        TreeNode* result = nullptr;
        if (root->val > val) result = searchBST(root->left, val);
        else if (root->val < val) result = searchBST(root->right, val);
        return result;
    }
};

迭代法：

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        while (root!=nullptr)
        {
            if (root->val > val) root = root->left;
            else if (root->val < val) root = root->right;
            else return root;
        }
        return nullptr;
    }
};

Leetcode 98-验证二叉搜索树

题目描述

https://leetcode.cn/problems/validate-binary-search-tree/description/

解题思路

要充分利用二叉搜索树的特性，要使用中序遍历（左中右），在这样的遍历顺序下，遍历的值是逐渐递增的。

要注意二叉搜索树不仅仅是中节点的值大于左节点小于右节点，同时需要保证中节点的值大于左子树的所有值，小于右子树的所有值。

class Solution {
public:
    TreeNode* pre = nullptr;
    bool isValidBST(TreeNode* root) {
        if (root == nullptr)return true;
        bool left = isValidBST(root->left);
        if (pre != nullptr && root->val <= pre->val)return false;
        else pre = root;
        bool right = isValidBST(root->right);
        return left && right;
    }
};