113. Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
solution:
本题意是从根节点到叶子节点找出所有相加之和等于sum的路径,并存在一条路径为一维数组。解决的方法和之前的I差不多,都是递归遍历左右节点。每次递归都减去相应节点的值,每次递归前将此节点的值存进vector容器内。当到达叶子节点时,如果sum值等于叶子节点的值,则这条路径满足条件,将这条路径即path添加到新的一个vector容器内paths。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void pathSum(vector<vector<int> > &paths, vector<int> &path, TreeNode *tree, int sum)
{
if(!tree) return;
path.push_back(tree->val);
if(!tree->left && !tree->right && tree->val==sum)
paths.push_back(path);
if(tree->left)
{
pathSum(paths,path,tree->left,sum-tree->val);
path.pop_back();
}
if(tree->right)
{
pathSum(paths,path,tree->right,sum-tree->val);
path.pop_back();
}
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<vector<int> > paths;
vector<int> path;
pathSum(paths,path,root,sum);
return paths;
}
};