113. Path Sum II

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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