算法打卡 Day22（二叉树）-最大二叉树 + 合并二叉树 + 二叉搜索树中的搜索
Leetcode 654-最大二叉树题目形貌
https://leetcode.cn/problems/maximum-binary-tree/description/
https://i-blog.csdnimg.cn/direct/259e7de7326f47b1949c2b5e6921d0b7.png#pic_left
解题思路
class Solution {
public:
TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
TreeNode* root = new TreeNode(nums);
if (nums.size() == 1)return root;
int maxIndex = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums > root->val) {
root->val = nums;
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
题目形貌
https://leetcode.cn/problems/merge-two-binary-trees/description/
https://i-blog.csdnimg.cn/direct/ae30ba2a369644e083e07a8f3dd92648.png#pic_left
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
题目形貌
https://leetcode.cn/problems/search-in-a-binary-search-tree/description/
https://i-blog.csdnimg.cn/direct/e500a97f52a94308a5f721ca6f5671fe.png#pic_left
解题思路
二叉搜索树：二叉搜索树自带序次，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
题目形貌
https://leetcode.cn/problems/validate-binary-search-tree/description/
https://i-blog.csdnimg.cn/direct/b48ce8f0f2b04d9e8859c3293f7af14b.png#pic_left
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），在如许的遍历序次下，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全部值，小于右子树的全部值。
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;
}
};
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