113. Path Sum II

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题目

 Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

解析

采用dfs和bfs的方法，很经典的方法，类似的题目很多都可以采用此方法，熟练掌握！

class Solution_113 { public:

void dfs(TreeNode* root,int cur_sum,int sum,vector<int> &vec ,vector<vector<int>> &vecs)
{
    if (!root)
    {
        return;
    }

    if (root->left==NULL&&root->right==NULL&&cur_sum==sum)
    {
        vecs.push_back(vec);
        return;
    }
    if (root->left)
    {
        vec.push_back(root->left->val);
        dfs(root->left, cur_sum + root->left->val, sum, vec, vecs);
        vec.pop_back();
    }
    if (root->right)
    {
        vec.push_back(root->right->val);
        dfs(root->right, cur_sum + root->right->val, sum, vec, vecs);
        vec.pop_back();
    }
    
    return;
}

vector<vector<int> > pathSum1(TreeNode *root, int sum) {
    vector<vector<int>> vecs;
    vector<int> vec;

    if (!root)
    {
        return vecs;
    }

    vec.push_back(root->val);
    dfs(root,root->val,sum,vec,vecs); //输入当前节点及其当前节点的和
    return vecs;
}

vector<vector<int> > pathSum(TreeNode *root, int sum) {

    vector<vector<int>> vecs;

    if (!root)
    {
        return vecs;
    }

    queue<TreeNode*> que;
    que.push(root);

    queue<vector<int>> path;
    path.push({ root->val }); 

    while (!que.empty())
    {
        TreeNode* temp;
        int size = que.size();

        for (int i = 0; i < size;i++)
        {
            temp = que.front();
            que.pop();

            vector<int>  vec= path.front();
            path.pop();

            if (temp->left==NULL&&temp->right==NULL&& accumulate(vec.begin(),vec.end(),0)==sum) //0 累加的初始值
            {
                vecs.push_back(vec);
            }
            if (temp->left)
            {
                que.push(temp->left);

                vector<int> var = vec;
                var.push_back(temp->left->val);
                path.push(var);
                //vec.pop_back();
            }
            if (temp->right)
            {
                que.push(temp->right);
                vector<int> var = vec;
                var.push_back(temp->right->val);
                path.push(var);
                //vec.pop_back();
            }
        }
    }
    return vecs;
}

};