leetcode: Path Sum II_binary search sum path-优快云博客

本文链接：https://blog.youkuaiyun.com/binary_search/article/details/16923159

本文介绍了一种算法，用于查找二叉树中所有从根节点到叶子节点的路径，其中路径上节点值的总和等于给定的数值。通过递归方式实现，详细展示了如何遍历树并收集符合条件的路径。

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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]
]

思路

树的遍历递归，注意vector中节点的顺序。这里我想令写一个函数，换成返回值是一个vector<deque<int> >应该会更好，这样就不用把顺序反过来了，直接push_front就好了。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
vector<vector<int> > pathSum(TreeNode *root, int sum) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		vector<vector<int> > res;
        if(root == NULL) return res;
		if(root->left == NULL && root->right == NULL)
		{
			if(sum == root->val)
			{
				vector<int> tmp;
				tmp.push_back(root->val);
				res.push_back(tmp);
			}
		}
		else
		{
			if(root->left !=NULL)
			{
				vector<vector<int> > resl=pathSum(root->left, sum - root->val);
				//if(resl.size() == 0) return res;//root->left == NULL or has NO path
				for(int i = 0; i < resl.size(); i ++)
				{
					if(resl[i].size() != 0)
					{
						vector<int> tmp;
						tmp.push_back(root->val);
						for(int j = 0; j < resl[i].size(); j++)
						{
							tmp.push_back(resl[i][j]);
						}
						res.push_back(tmp);
					}
				}
			}

			if(root->right !=NULL)
			{
				vector<vector<int> > resr=pathSum(root->right, sum - root->val);
				//if(resl.size() == 0) return res;//root->left == NULL or has NO path
				for(int i = 0; i < resr.size(); i ++)
				{
					if(resr[i].size() != 0)
					{
						vector<int> tmp;
						tmp.push_back(root->val);
						for(int j = 0; j < resr[i].size(); j++)
						{
							tmp.push_back(resr[i][j]);
						}
						res.push_back(tmp);
					}
				}
			}
		}
		return res;
    }
};