代码随想录算法训练营第18天 | 513.找树左下角的值、112. 路径总和、113.路径总和ii、106.从中序与后序遍历序列构造二叉树、105.从前序与中序遍历序列构造二叉树

发布于 2022-12-03  499 次阅读


513. 找树左下角的值 - 力扣(LeetCode)

思路:

  • 迭代法:层序遍历二叉树,每次都将该层第一个值赋给ans,最后返回ans
  • 递归法:前序遍历二叉树,遇到左右子树都为nullptr则更新最大深度和结果值,每次递归进行前序遍历,即先递归左子树,再递归右子树,传递的深度值为depth + 1

我的AC代码

层序遍历迭代法

//时间复杂度O(n),空间复杂度O(n)
class Solution {
public:
    int findBottomLeftValue(TreeNode* root) {
        queue<TreeNode*> que;
        int ans = root->val;
        que.push(root);
        while(!que.empty()) {
            int qsize = que.size();
            int qcnt = qsize;
            while(qsize--) {
                TreeNode* node = que.front();
                if(qcnt - 1 == qsize) {
                    ans = node->val;
                }
                que.pop();
                if(node->left) {
                    que.push(node->left);
                }
                if(node->right) {
                    que.push(node->right);
                }
            }
        }
        return ans;
    }
};

标准答案

递归(前序遍历)

//时间复杂度O(n),空间复杂度O(n)
class Solution {
public:
    int maxDepth = INT_MIN;
    int result;
    void traversal(TreeNode* root, int depth) {
        if (root->left == NULL && root->right == NULL) {
            if (depth > maxDepth) {
                maxDepth = depth;
                result = root->val;
            }
            return;
        }
        if (root->left) {
            depth++;
            traversal(root->left, depth);
            depth--; // 回溯
        }
        if (root->right) {
            depth++;
            traversal(root->right, depth);
            depth--; // 回溯
        }
        return;
    }
    int findBottomLeftValue(TreeNode* root) {
        traversal(root, 0);
        return result;
    }
};

迭代(层序遍历)

//时间复杂度O(n),空间复杂度O(n)
class Solution {
public:
    int findBottomLeftValue(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        int result = 0;
        while (!que.empty()) {
            int size = que.size();
            for (int i = 0; i < size; i++) {
                TreeNode* node = que.front();
                que.pop();
                if (i == 0) result = node->val; // 记录最后一行第一个元素
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
        }
        return result;
    }
};

112. 路径总和 - 力扣(LeetCode)

思路:递归法前序遍历二叉树,记录下所有的路径,每遇到一个节点就将节点值加上,如果遍历到的节点没有左右孩子则此时为路径末尾,判断总和是否与目标值相等,相等则将ans赋值为true

我的AC代码

//时间复杂度O(n),空间复杂度O(n)
//递归法(前序遍历)
class Solution {
public:
    bool ans = false;
    int target;
    void findpath(TreeNode* node, int sum) {
        if(!node->left && !node->right) {
            sum += node->val;
            if(sum == target) {
                ans = true;
            }
            return;
        }
        sum += node->val;
        if(node->left) {
            findpath(node->left, sum);
        }
        if(node->right) {
            findpath(node->right, sum);
        }
    }
    bool hasPathSum(TreeNode* root, int targetSum) {
        if(root == nullptr) {
            return false;
        }
        target = targetSum;
        int sum = 0;
        findpath(root, sum);
        return ans;
    }
};

标准答案

递归(前序遍历)完整版

//时间复杂度O(n),空间复杂度O(n)
class Solution {
private:
    bool traversal(TreeNode* cur, int count) {
        if (!cur->left && !cur->right && count == 0) return true; // 遇到叶子节点,并且计数为0
        if (!cur->left && !cur->right) return false; // 遇到叶子节点直接返回

        if (cur->left) { // 左
            count -= cur->left->val; // 递归,处理节点;
            if (traversal(cur->left, count)) return true;
            count += cur->left->val; // 回溯,撤销处理结果
        }
        if (cur->right) { // 右
            count -= cur->right->val; // 递归,处理节点;
            if (traversal(cur->right, count)) return true;
            count += cur->right->val; // 回溯,撤销处理结果
        }
        return false;
    }

public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (root == NULL) return false;
        return traversal(root, sum - root->val);
    }
};

递归(前序遍历)精简版

//时间复杂度O(n),空间复杂度O(n)
class solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (root == null) return false;
        if (!root->left && !root->right && sum == root->val) {
            return true;
        }
        return haspathsum(root->left, sum - root->val) || haspathsum(root->right, sum - root->val);
    }
};

迭代法(前序遍历)

//时间复杂度O(n),空间复杂度O(n)
class solution {
public:
    bool haspathsum(TreeNode* root, int sum) {
        if (root == null) return false;
        // 此时栈里要放的是pair<节点指针,路径数值>
        stack<pair<TreeNode*, int>> st;
        st.push(pair<TreeNode*, int>(root, root->val));
        while (!st.empty()) {
            pair<TreeNode*, int> node = st.top();
            st.pop();
            // 如果该节点是叶子节点了,同时该节点的路径数值等于sum,那么就返回true
            if (!node.first->left && !node.first->right && sum == node.second) return true;

            // 右节点,压进去一个节点的时候,将该节点的路径数值也记录下来
            if (node.first->right) {
                st.push(pair<TreeNode*, int>(node.first->right, node.second + node.first->right->val));
            }

            // 左节点,压进去一个节点的时候,将该节点的路径数值也记录下来
            if (node.first->left) {
                st.push(pair<TreeNode*, int>(node.first->left, node.second + node.first->left->val));
            }
        }
        return false;
    }
};

113. 路径总和 II - 力扣(LeetCode)

思路:递归法前序遍历二叉树,记录下所有的路径,每遇到一个节点就将节点值加上,如果遍历到的节点没有左右孩子则此时为路径末尾,判断总和是否与目标值相等,相等则将ans赋值为true并将路径添加到结果数组

我的AC代码

//时间复杂度O(n),空间复杂度O(n)
//递归法(前序遍历)
class Solution {
public:
    int target;
    void get_path(vector<vector<int>>& ans,int sum, vector<int>& path, TreeNode*node) {
        sum += node->val;
        path.push_back(node->val);
        if(!node->left && !node->right) {
            if(sum == target) {
                ans.push_back(path);
            }
        }
        if(node->left) {
            get_path(ans, sum, path, node->left);
            path.pop_back();
        }
        if(node->right) {
            get_path(ans, sum, path, node->right);
            path.pop_back();
        }
    }
    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        vector<vector<int>> ans;
        vector<int> path;
        target = targetSum;
        if(root == nullptr) {
            return ans;
        }
        int sum = 0;
        get_path(ans, sum, path, root);
        return ans;
    }
};

标准答案

//时间复杂度O(n),空间复杂度O(n)
//递归法(前序遍历)
class solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    // 递归函数不需要返回值,因为我们要遍历整个树
    void traversal(treenode* cur, int count) {
        if (!cur->left && !cur->right && count == 0) { // 遇到了叶子节点且找到了和为sum的路径
            result.push_back(path);
            return;
        }

        if (!cur->left && !cur->right) return ; // 遇到叶子节点而没有找到合适的边,直接返回

        if (cur->left) { // 左 (空节点不遍历)
            path.push_back(cur->left->val);
            count -= cur->left->val;
            traversal(cur->left, count);    // 递归
            count += cur->left->val;        // 回溯
            path.pop_back();                // 回溯
        }
        if (cur->right) { // 右 (空节点不遍历)
            path.push_back(cur->right->val);
            count -= cur->right->val;
            traversal(cur->right, count);   // 递归
            count += cur->right->val;       // 回溯
            path.pop_back();                // 回溯
        }
        return ;
    }

public:
    vector<vector<int>> pathsum(treenode* root, int sum) {
        result.clear();
        path.clear();
        if (root == null) return result;
        path.push_back(root->val); // 把根节点放进路径
        traversal(root, sum - root->val);
        return result;
    }
};

106. 从中序与后序遍历序列构造二叉树 - 力扣(LeetCode)

思路:想要用中序和后序遍历序列确定一个唯一的二叉树,就要每次以后序数组的最后一个元素为切割点,将中序数组分割成左右两份,再根据中序数组的分割情况来切割后序数组,如此反复,每次的切割点就是中间的节点

借用carl哥的图片帮助理解一下

我的AC代码

递归法(每次都定义新的vector)

//时间复杂度O(n),空间复杂度O(n2)
class Solution {
public:
    TreeNode* build(vector<int>& inorder, vector<int>& postorder) {
        int psize = postorder.size();
        if(psize == 0) {
            return nullptr;
        }
        TreeNode* root = new TreeNode(postorder[psize - 1]);
        if(psize == 1) {
            return root;
        }
        int sign = 0;
        for(int i = 0; i < psize; ++i) {
            if(inorder[i] == postorder[psize - 1]) {
                sign = i;
                break;
            }
        }

        vector<int> leftinorder(inorder.begin(), inorder.begin() + sign);
        vector<int> rightinorder(inorder.begin() + sign + 1, inorder.end());    
        postorder.resize(psize - 1);
        vector<int> leftpostorder(postorder.begin(), postorder.begin() + leftinorder.size());
        vector<int> rightpostorder(postorder.begin() + leftinorder.size(), postorder.end());
        root->left = build(leftinorder, leftpostorder);
        root->right = build(rightinorder, rightpostorder);
        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int isize = inorder.size();
        if(isize == 1) {
            TreeNode* tmp = new TreeNode(inorder[0]);
            return tmp;
        }
        return build(inorder, postorder);
    }
};

递归法(不重新定义新的vector而是建立索引,大大降低空间复杂度)

//时间复杂度O(n),空间复杂度O(n)
class Solution {
public:
    TreeNode* build(vector<int>& inorder, int inbegin, int inend, vector<int>& postorder, int pobegin, int poend) {
        int psize = poend - pobegin;
        if(psize == 0) {
            return nullptr;
        }
        TreeNode* root = new TreeNode(postorder[poend - 1]);
        if(psize == 1) {
            return root;
        }
        int sign;
        for(int i = inbegin; i < inend; ++i) {
            if(inorder[i] == postorder[poend - 1]) {
                sign = i;
            }
        }
        int leftinbegin = inbegin;
        int leftinend = sign;

        int rightinbegin =  sign + 1;
        int rightinend = inend;

        int leftpobegin = pobegin;
        int leftpoend = pobegin + sign - inbegin;

        int rightpobegin = pobegin + sign - inbegin;
        int rightpoend = poend - 1;

        root->left = build(inorder, leftinbegin, leftinend, postorder, leftpobegin, leftpoend);
        root->right = build(inorder, rightinbegin, rightinend, postorder, rightpobegin, rightpoend);

        return root;

    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int isize = inorder.size();
        int psize = postorder.size();
        return build(inorder, 0, isize, postorder, 0, psize);

    }
};

标准答案

递归法(每次都定义新的vector)

//时间复杂度O(n),空间复杂度O(n)
class Solution {
private:
    TreeNode* traversal (vector<int>& inorder, vector<int>& postorder) {
        if (postorder.size() == 0) return NULL;

        // 后序遍历数组最后一个元素,就是当前的中间节点
        int rootValue = postorder[postorder.size() - 1];
        TreeNode* root = new TreeNode(rootValue);

        // 叶子节点
        if (postorder.size() == 1) return root;

        // 找到中序遍历的切割点
        int delimiterIndex;
        for (delimiterIndex = 0; delimiterIndex < inorder.size(); delimiterIndex++) {
            if (inorder[delimiterIndex] == rootValue) break;
        }

        // 切割中序数组
        // 左闭右开区间:[0, delimiterIndex)
        vector<int> leftInorder(inorder.begin(), inorder.begin() + delimiterIndex);
        // [delimiterIndex + 1, end)
        vector<int> rightInorder(inorder.begin() + delimiterIndex + 1, inorder.end() );

        // postorder 舍弃末尾元素
        postorder.resize(postorder.size() - 1);

        // 切割后序数组
        // 依然左闭右开,注意这里使用了左中序数组大小作为切割点
        // [0, leftInorder.size)
        vector<int> leftPostorder(postorder.begin(), postorder.begin() + leftInorder.size());
        // [leftInorder.size(), end)
        vector<int> rightPostorder(postorder.begin() + leftInorder.size(), postorder.end());

        root->left = traversal(leftInorder, leftPostorder);
        root->right = traversal(rightInorder, rightPostorder);

        return root;
    }
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if (inorder.size() == 0 || postorder.size() == 0) return NULL;
        return traversal(inorder, postorder);
    }
};

递归法(不重新定义新的vector而是建立索引,大大降低空间复杂度)

//时间复杂度O(n),空间复杂度O(n)
class Solution {
private:
    // 中序区间:[inorderBegin, inorderEnd),后序区间[postorderBegin, postorderEnd)
    TreeNode* traversal (vector<int>& inorder, int inorderBegin, int inorderEnd, vector<int>& postorder, int postorderBegin, int postorderEnd) {
        if (postorderBegin == postorderEnd) return NULL;

        int rootValue = postorder[postorderEnd - 1];
        TreeNode* root = new TreeNode(rootValue);

        if (postorderEnd - postorderBegin == 1) return root;

        int delimiterIndex;
        for (delimiterIndex = inorderBegin; delimiterIndex < inorderEnd; delimiterIndex++) {
            if (inorder[delimiterIndex] == rootValue) break;
        }
        // 切割中序数组
        // 左中序区间,左闭右开[leftInorderBegin, leftInorderEnd)
        int leftInorderBegin = inorderBegin;
        int leftInorderEnd = delimiterIndex;
        // 右中序区间,左闭右开[rightInorderBegin, rightInorderEnd)
        int rightInorderBegin = delimiterIndex + 1;
        int rightInorderEnd = inorderEnd;

        // 切割后序数组
        // 左后序区间,左闭右开[leftPostorderBegin, leftPostorderEnd)
        int leftPostorderBegin =  postorderBegin;
        int leftPostorderEnd = postorderBegin + delimiterIndex - inorderBegin; // 终止位置是 需要加上 中序区间的大小size
        // 右后序区间,左闭右开[rightPostorderBegin, rightPostorderEnd)
        int rightPostorderBegin = postorderBegin + (delimiterIndex - inorderBegin);
        int rightPostorderEnd = postorderEnd - 1; // 排除最后一个元素,已经作为节点了

        root->left = traversal(inorder, leftInorderBegin, leftInorderEnd,  postorder, leftPostorderBegin, leftPostorderEnd);
        root->right = traversal(inorder, rightInorderBegin, rightInorderEnd, postorder, rightPostorderBegin, rightPostorderEnd);

        return root;
    }
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if (inorder.size() == 0 || postorder.size() == 0) return NULL;
        // 左闭右开的原则
        return traversal(inorder, 0, inorder.size(), postorder, 0, postorder.size());
    }
};

105. 从前序与中序遍历序列构造二叉树 - 力扣(LeetCode)

思路:与上一题一致,上一题的节点是后序遍历序列的最后一个,而这一题的节点是前序遍历序列的第一个

我的AC代码

// 时间复杂度O(n),空间复杂度O(n)
// 使用了建立下标索引的方法来优化空间复杂度
class Solution {
public:
    TreeNode* build(vector<int>& preorder, int preBegin, int preEnd, vector<int>& inorder, int inBegin, int inEnd) {
        int psize = preEnd - preBegin;
        if(psize == 0) {
            return nullptr;
        }
        TreeNode* root = new TreeNode(preorder[preBegin]);
        if(psize == 1) {
            return root;
        }
        int sign;
        for(int i = inBegin; i < inEnd; ++i) {
            if(inorder[i] == preorder[preBegin]) {
                sign = i;
            }
        }

        int leftInBegin = inBegin;
        int leftInEnd = sign;

        int rightInBegin = sign + 1;
        int rightInEnd = inEnd;

        int leftPreBegin = preBegin + 1;
        int leftPreEnd = preBegin + sign - inBegin + 1;

        int rightPreBegin = preBegin + sign - inBegin + 1;
        int rightPreEnd = preEnd;

        root->left = build(preorder, leftPreBegin, leftPreEnd, inorder, leftInBegin, leftInEnd);
        root->right = build(preorder, rightPreBegin, rightPreEnd, inorder, rightInBegin, rightInEnd);        

        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return build(preorder, 0, preorder.size(), inorder, 0, inorder.size());
    }
};

标准答案

// 时间复杂度O(n),空间复杂度O(n)
// 使用了建立下标索引的方法来优化空间复杂度
class Solution {
private:
        TreeNode* traversal (vector<int>& inorder, int inorderBegin, int inorderEnd, vector<int>& preorder, int preorderBegin, int preorderEnd) {
        if (preorderBegin == preorderEnd) return NULL;

        int rootValue = preorder[preorderBegin]; // 注意用preorderBegin 不要用0
        TreeNode* root = new TreeNode(rootValue);

        if (preorderEnd - preorderBegin == 1) return root;

        int delimiterIndex;
        for (delimiterIndex = inorderBegin; delimiterIndex < inorderEnd; delimiterIndex++) {
            if (inorder[delimiterIndex] == rootValue) break;
        }
        // 切割中序数组
        // 中序左区间,左闭右开[leftInorderBegin, leftInorderEnd)
        int leftInorderBegin = inorderBegin;
        int leftInorderEnd = delimiterIndex;
        // 中序右区间,左闭右开[rightInorderBegin, rightInorderEnd)
        int rightInorderBegin = delimiterIndex + 1;
        int rightInorderEnd = inorderEnd;

        // 切割前序数组
        // 前序左区间,左闭右开[leftPreorderBegin, leftPreorderEnd)
        int leftPreorderBegin =  preorderBegin + 1;
        int leftPreorderEnd = preorderBegin + 1 + delimiterIndex - inorderBegin; // 终止位置是起始位置加上中序左区间的大小size
        // 前序右区间, 左闭右开[rightPreorderBegin, rightPreorderEnd)
        int rightPreorderBegin = preorderBegin + 1 + (delimiterIndex - inorderBegin);
        int rightPreorderEnd = preorderEnd;

        root->left = traversal(inorder, leftInorderBegin, leftInorderEnd,  preorder, leftPreorderBegin, leftPreorderEnd);
        root->right = traversal(inorder, rightInorderBegin, rightInorderEnd, preorder, rightPreorderBegin, rightPreorderEnd);

        return root;
    }

public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        if (inorder.size() == 0 || preorder.size() == 0) return NULL;

        // 参数坚持左闭右开的原则
        return traversal(inorder, 0, inorder.size(), preorder, 0, preorder.size());
    }
};