LeetCode Path Sum II路径和II

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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]]

跟PathSumI一样

http://blog.csdn.net/kenden23/article/details/14223191

检查一颗二叉树是否有和某数相等的路径。和前面的最少深度二叉树思路差不多。

1 每进入一层，如果该层跟节点不为空，那么就加上这个值。

2 如果到了叶子节点，那么就与这个特定数比较，相等就说明存在这样的路径。

当然可以不是用递归优化一点算法吧。是用循环的话就可以在找到值的时候直接返回。递归也可以增加判断，找到的时候每层加判断迅速返回，不过感觉效果也高不了多少。

/** * 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) {vector<int> onePath; vector<vector<int> > path;paSum(root, onePath, path, sum);return path;}void paSum(TreeNode *node,vector<int> &onePath, vector<vector<int> > &path,int overall, int levelNum = 0){if(node == nullptr) return;onePath.push_back(node->val);levelNum += node->val;paSum(node->left, onePath, path, overall, levelNum);paSum(node->right, onePath, path, overall, levelNum);if(node->left==nullptr && node->right==nullptr && overall==levelNum)path.push_back(onePath);onePath.pop_back();}};