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113. Path Sum II

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;
    }
};

题目来源

• 113. Path Sum II