Code Caprice Algorithm Training Camp Day 18|513. Find the value in the lower left corner of the tree| Path sum| Construct a binary tree from inorder and postorder traversal sequences

513. Find the value in the lower left corner of the tree

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    
    
public:
    int findBottomLeftValue(TreeNode* root) {
    
    
        queue<TreeNode*>que;
        int res;
        que.push(root);
        while(!que.empty()){
    
    
            int size=que.size();
            TreeNode*tmp= que.front();
            res=tmp->val;
            for(int i=0;i<size;i++){
    
    
                tmp= que.front();
                que.pop();
                if(tmp->left)que.push(tmp->left);
                if(tmp->right)que.push(tmp->right); 
            }
        }
        return res;
    }
};

path sum

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    
    
    int res=0;
    void backing(int a,TreeNode*root,int targetSum){
    
    
        if(root->left==nullptr&&root->right==nullptr&&a==targetSum){
    
    
            res=1;
            return ;
        }
        if(root->left==nullptr&&root->right==nullptr){
    
    
            return ;
        }
        if(root->left!=nullptr){
    
    
            backing(a+root->left->val,root->left,targetSum);
        }
        if(root->right!=nullptr){
    
    
            backing(a+root->right->val,root->right,targetSum);
        }

    }
public:
    bool hasPathSum(TreeNode* root, int targetSum) {
    
    
        if(root==nullptr)return 0;
        backing(root->val,root,targetSum);

        return res;

    }
};

Construct binary tree from inorder and postorder traversal sequences

 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);
    }
};