算法训练1-day15-二叉树

今天是二叉树

1. 654. 最大二叉树

此题还有单调栈的O(n)解法

AC代码：

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        return _constructMaximumBinaryTree(nums, 0, nums.size() - 1);
    }

    TreeNode* _constructMaximumBinaryTree(vector<int>& nums, int left, int right) {
        if (left > right) return nullptr;

        int maxPos = left;
        int maxVal = nums[left];
        for (int i = left + 1; i <= right; ++i) {
            if (nums[i] > nums[maxPos]) {
                maxVal = nums[i];
                maxPos = i;
            }
        }
        TreeNode* root = new TreeNode(maxVal);

        root->left = _constructMaximumBinaryTree(nums, left, maxPos - 1);
        root->right = _constructMaximumBinaryTree(nums, maxPos + 1, right);

        return root;
    }
};

2. 617. 合并二叉树

AC代码：

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        if(root1 == nullptr) return root2;
        else if(root2 == nullptr) return root1;

        root1->val += root2->val;

        root1->left = mergeTrees(root1->left, root2->left);
        root1->right = mergeTrees(root1->right, root2->right);

        return root1;
    }
};

3. 700. 二叉搜索树中的搜索

AC代码：

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if(root == nullptr) return nullptr;
        if(root->val == val) return root;

        if(root->val < val){
            return searchBST(root->right, val);
        }

        return searchBST(root->left, val);
    }
};

4. 98. 验证二叉搜索树

class Solution {
public:
    bool isValidBST(TreeNode* root) {
        vector<int> nums;
        inorder(nums, root);
        for (int i = 1; i < nums.size(); ++i) {
            if (nums[i - 1] >= nums[i]) return false;
        }

        return true;
    }

    void inorder(vector<int>& nums, TreeNode* root) {
        if (root == nullptr) return;

        inorder(nums, root->left);
        nums.push_back(root->val);
        inorder(nums, root->right);
    }
};

//除了使用数组来记录中序遍历的结果外
//我们还可以使用一个pre指针来记录前一个节点
//然后比较当前节点的值是否大于pre的值
//大于，则是正确的；反之，就不是二叉搜索树
class Solution {
public:
    TreeNode* pre = NULL; // 用来记录前一个节点
    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; // 记录前一个节点

        bool right = isValidBST(root->right);
        return left && right;
    }
};