[leetcode] Path Sum II

最新推荐文章于 2017-06-29 10:05:00 发布

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最新推荐文章于 2017-06-29 10:05:00 发布

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分类专栏：算法题

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本文介绍了一种算法，用于在给定的二叉树中找到所有从根节点到叶子节点的路径，使得这些路径上的元素之和等于指定的数值。通过实例演示了如何实现这一算法。

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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,

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]
]

class Solution {
public:
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		vector<vector<int> > ret;
		vector<int> path;
		_pathSum(root,sum,path,ret);
		return ret;
	}
	void _pathSum(TreeNode *root, int sum, vector<int> &path, vector<vector<int> >& ret){
		if(root==NULL)
			return;
		if(!root->left && !root->right && root->val==sum){
			path.push_back(root->val);
			ret.push_back(path);
            path.pop_back();
		}
		if(!root->left && !root->right && root->val!=sum){
			return;
		}
		path.push_back(root->val);
		if(root->left)
			_pathSum(root->left,sum-root->val,path,ret);
        if(root->right)
			_pathSum(root->right,sum-root->val,path,ret);
        path.pop_back();

	}
};