代码随想录算法训练营第21天 | 第六章 二叉树part08

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今日记录

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669.修剪二叉树

Leetcode链接

关键点在于root->val 如果在 [ low, high]以外时怎么正确的删除节点

class Solution {
public:
    TreeNode* trimBST(TreeNode* root, int low, int high) {
        if (root == NULL)
            return NULL;
        if (root->val < low) {
            TreeNode* rightnode = trimBST(root->right, low, high);
            return rightnode;
        }
        if (root->val > high) {
            TreeNode* leftnode = trimBST(root->left, low, high);
            return leftnode;
        }
        root->left = trimBST(root->left, low, high);
        root->right = trimBST(root->right, low, high);
        return root;
    }
};

108.将有序数组转换为二叉搜索树

Leetcode链接

class Solution {
private:
    TreeNode* traversal(vector<int>& nums, int left, int right) {
        if (left > right)
            return NULL;
        int mid = left + ((right - left) / 2); // 防止越界
        TreeNode* root = new TreeNode(nums[mid]);
        root->left = traversal(nums, left, mid - 1);
        root->right = traversal(nums, mid + 1, right);
        return root;
    }
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        TreeNode* root = traversal(nums, 0, nums.size() - 1);
        return root;
    }
};

538. 把二叉搜索树转换为累加树

Leetcode链接

class Solution {
private:
    int pre = 0;
    void traversal(TreeNode* cur) {
        if (cur == NULL)
            return;
        traversal(cur->right); // 右
        cur->val += pre;
        pre = cur->val;
        traversal(cur->left);
    }

public:
    TreeNode* convertBST(TreeNode* root) {
        pre = 0;
        traversal(root);
        return root;
    }
};

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总结