问题描述:
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] ]
问题分析:
如题所述,很容易可以想到使用深度优先搜索遍历二叉树,然后维护一个数组,用来保存从根节点到当前节点所经过的路径的节点值,当遍历到叶子节点时,对访问路径所经过的节点的值求和,若与欲求的目标值相等,则为目标路径,否则返回,尝试下一条路径。
具体实现:
/**
* 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:
vector> pathSum(TreeNode* root, int sum) {
vector> res;
vector list;
searchSum(res,root,list,0,sum);
}
void searchSum(vector> &res, TreeNode* root, vector list, int val, int sumTarget)
{
if(res==NULL)
{
if(val==sumTarget)
res.pushback(list);
return;
}
list.pushback(root->val);
val = val+root->val;
searchSum(res, root->left,list,val,sumTarget);
searchSum(res, root->right,list,val,sumTarget);
}
};