513.找树左下角的值

力扣题目链接

int find(TreeNode* root) {
        queue<TreeNode*> que;
        int res;
        if (root != NULL) {
            que.push(root);
        }
        // 层次遍历
        while (!que.empty()) {
            int size = que.size();
            for (int i = 0; i < size; ++i) {
                TreeNode* node = que.front();
                que.pop();
                // 每次外层while循环即是一层节点，其中i = 0为该层最左边的节点，每一层都记录最左侧节点的值，即获得题目要求的最深处的最左值。
                if (i == 0) {
                    res = node->val;
                }
                if (node->left) {
                    que.push(node->left);
                }
                if (node->right) {
                    que.push(node->right);
                }
            }
        }
        return res;
    }

    int findBottomLeftValue(TreeNode* root) {
        int result = find(root);
        return result;
    }     

    // 递归法
    // 全局变量
    int maxDepth = INT16_MIN;
    int result;
    // 确定函数返回值和函数参数
    void traversal(TreeNode* root, int depth) {
        // 确定终止条件
        // 叶子节点
        if (root->left == NULL && root->right == NULL) {
            // 更新深度与赋值
            // 因为是根左右，所以只有最左边的值进入下逻辑
            if (depth > maxDepth) {
                maxDepth = depth;
                // 最左边的值
                result = root->val;
            }
        }
        // 确定单层循环条件
        if (root->left) {
            depth++;
            traversal(root->left, depth);
            depth--;
        }
        if (root->right) {
            depth++;
            traversal(root->right, depth);
            depth--;
        }
    }
    int findBottomLeftValue1(TreeNode* root) {
        traversal(root, 0);
        return result;
    } 

112. 路径总和

力扣题目链接

// 递归法
    bool traversal(TreeNode* cur, int count) {
        // 遇到叶子节点并且计数为零
        if (!cur->left && !cur->right && count == 0) {
            return true;
        }
        // 遇到叶子节点直接返回
        if (!cur->left && !cur->right) {
            return false;
        }
        if (cur->left) {
            count -= cur->left->val;
            if (traversal(cur->left, count)) {
                return true;
            }
            count += cur->left->val;
        }
        if (cur->right) {
            count -= cur->right->val;
            if (traversal(cur->right, count)) {
                return true;
            }
            count += cur->right->val;
        }
        return false;
    }

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

    // 栈模拟的前序遍历 迭代法
    bool haspathsum(TreeNode* root, int sum) {
        if (root == NULL) {
            return false;
        }
        stack<pair<TreeNode*, int>> st;
        st.push(pair<TreeNode*, int>(root, root->val));
        while (!st.empty()) {
            pair<TreeNode*, int> node = st.top();
            st.pop();
            if (!node.first->left && !node.first->right && sum == node.second) {
                return true;
            }
            if (node.first->right) {
                st.push(pair<TreeNode*, int> (node.first->right, node.second + node.first->right->val));
            }
            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

力扣题目链接

vector<vector<int>> result;
    vector<int> path;
    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;
        }

        if (cur->left) {
            path.push_back(cur->left->val);
            count -= cur->left->val;
            traversal(cur->left, count);
            count += cur->left->val;
            path.pop_back();
        }

        if (cur->right) { 
            path.push_back(cur->right->val);
            count -= cur->right->val;
            traversal(cur->right, count);   
            count += cur->right->val;       
            path.pop_back();               
        }
        return ;
    }

public:
    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        result.clear();
        path.clear();
        if (root == NULL) {
            return result;
        }
        path.push_back(root->val);
        traversal(root, targetSum - root->val);
        return result;
    }

106.从中序与后序遍历序列构造二叉树

力扣题目链接

// 中序遍历+后序遍历，构建二叉树,多打日志测试，不要头脑风暴
    TreeNode* traversal(vector<int>&inorder, vector<int>& postorder) {
        // 为空的情况
        if (postorder.size() == 0) {
            return NULL;
        }
        // 找到根节点，后序遍历的最后一个节点
        int rootVal = postorder[postorder.size() - 1];
        TreeNode* root = new TreeNode(rootVal);
        // 遇到叶子节点，返回点
        if (postorder.size() == 1) {
            return root;
        }
        // 找切割点，先切中序再切后序   
        int delimiterIndex;
        for (delimiterIndex = 0; delimiterIndex < inorder.size(); delimiterIndex++) {
          if (inorder[delimiterIndex] == rootVal) {
            break;
          }  
        }
        // 切割中序，左闭右开区间
        vector<int> leftInorder(inorder.begin(), inorder.begin() + delimiterIndex);
        vector<int> rightInorder(inorder.begin() + delimiterIndex + 1, inorder.end());
        // 切割后序，左闭右开区间
        postorder.resize(postorder.size() - 1);
        vector<int> leftPostorder(postorder.begin(), postorder.begin() + leftInorder.size());
        vector<int> rightPostorder(postorder.begin() + leftInorder.size(), postorder.end());
        // 递归
        root->left = traversal(leftInorder, leftPostorder);
        root->right = traversal(rightInorder, rightPostorder);

        // 出口
        return root;

    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return traversal(inorder, postorder);
    }

    // 改良版
    TreeNode* traversal(vector<int>& inorder, int inorderBegin, int inorderEnd, vector<int>& postorder, int postorderBegin, int postorderEnd) {
        if (postorderBegin == postorderEnd) {
            return NULL;
        }
        int rootVal = postorder[postorderEnd - 1];
        TreeNode* root = new TreeNode(rootVal);

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

        int delimiterIndex;
        for (delimiterIndex = inorderBegin; delimiterIndex < inorderEnd; ++delimiterIndex) {
            if (inorder[delimiterIndex] == rootVal) {
                break;
            }
        }

        int leftInBegin = inorderBegin;
        int leftInEnd = delimiterIndex;

        int rightInBegin = delimiterIndex + 1;
        int rightInEnd = inorderEnd;

        int leftPostBegin = postorderBegin;
        int leftPostEnd = postorderBegin + delimiterIndex - inorderBegin;

        int rightPostBegin = postorderBegin + delimiterIndex - inorderBegin;
        int rightPostEnd = postorderEnd - 1; // 排除最后一个元素，根节点

        root->left = traversal(inorder, leftInBegin, leftInEnd, postorder, leftPostBegin, leftPostEnd);
        root->right = traversal(inorder, rightInBegin, rightInEnd, postorder, rightPostBegin, rightPostEnd);

        return root;
    }

    TreeNode* buildTree1(vector<int>& inorder, vector<int>& postorder) {
        if (inorder.size() == 0 || postorder.size() == 0) {
            return NULL;
        }
        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

力扣题目链接

TreeNode* traversal(vector<int>& preorder, vector<int>& inorder) {
        if (preorder.size() == 0 || inorder.size() == 0) {
            return NULL;
        }
        int rootVal = *preorder.begin();
        TreeNode* root = new TreeNode(rootVal);

        if (preorder.size() == 1) {
            return root;
        }

        int delimiterIndex = -1;
        for (int i = 0; i < inorder.size(); ++i) {
            if (inorder[i] == rootVal) {
                delimiterIndex = i;
                break;
            }
        }
        
        vector<int> leftInorder(inorder.begin(), inorder.begin() + delimiterIndex);
        vector<int> rightInorder(inorder.begin() + delimiterIndex + 1, inorder.end());

        vector<int> leftPreorder(preorder.begin() + 1, preorder.begin() + delimiterIndex + 1);
        vector<int> rightPreorder(preorder.begin() + delimiterIndex + 1, preorder.end());


        root->left = traversal(leftPreorder, leftInorder);
        root->right = traversal(rightPreorder, rightInorder);

        return root;
    }

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return traversal(preorder, inorder);
    }