Path Sum && II

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool hasPathSum(struct TreeNode* root, int sum) {
    bool left, right;
    if(root == NULL)
        return false;
    else if(root->val == sum && root->left == NULL && root->right == NULL)
        return true;
    else
        {
            left = hasPathSum(root->left, sum - root->val);
            right = hasPathSum(root->right, sum - root->val);
            return (left || right);
        }
}
  • 关于DFS

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]
]

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> pathSum(TreeNode* root, int sum) {
13         vector<vector<int>> result;
14         vector<int> bkup;
15         int index = 0; // index of result
16         int addSum = 0;
17         // go over the tree, keep a addSum, 
18         // if addSum > sum, return 
19         DFS(sum, root, bkup, addSum, result, index);
20         return result;
21     }
22 private:
23     void DFS(int sum, TreeNode* node, vector<int> bkup, int addSum, vector<vector<int>> &result, int &index){
24         if(node == NULL)
25             return;
26         bkup.push_back(node->val);
27         addSum += node->val;
28         if (addSum == sum && node->left == NULL & node->right == NULL) {        
29             result.push_back(bkup);
30         }
31         else  {
32             DFS(sum, node->left, bkup, addSum, result, index);
33             DFS(sum, node->right, bkup, addSum, result, index);
34         }
35         return;
36     }
37 };
  • 主要是C++中vector的使用，直接push_back

 

posted @ 2015-10-11 21:38  dylqt  阅读(138)  评论(0)  编辑  收藏  举报