LeetCode:Path Sum I II

LeetCode:Path Sum

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.

递归求解，代码如下：

 /**
  * Definition for binary tree
  * struct TreeNode {
  * int val;
  * TreeNode *left;
  * TreeNode *right;
  * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     bool hasPathSum(TreeNode *root, int sum) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         if(root == NULL)return false;
         if(root->left == NULL && root->right == NULL && root->val == sum)
             return true;
         if(root->left && hasPathSum(root->left, sum - root->val))
             return true;
         if(root->right && hasPathSum(root->right, sum - root->val))
             return true;
         return false;
     }
 };

---

LeetCode: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.

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]
]                                                                       本文地址

同理也是递归求解，只是要保存当前的结果，并且每次递归出来后要恢复递归前的结果，每当递归到叶子节点时就把当前结果保存下来。

 /**
  * 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) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         vector<vector<int> > res;
         if(root == NULL)return res;
         vector<int> curres;
         curres.push_back(root->val);
         pathSumRecur(root, sum, curres, res);
         return res;
     }
     void pathSumRecur(TreeNode *root, int sum, vector<int> &curres, vector<vector<int> >&res)
     {
         if(root->left == NULL && root->right == NULL && root->val == sum)
         {
             res.push_back(curres);
             return;
         }
         if(root->left)
         {
             curres.push_back(root->left->val);
             pathSumRecur(root->left, sum - root->val, curres, res);
             curres.pop_back();
         }
         if(root->right)
         {
             curres.push_back(root->right->val);
             pathSumRecur(root->right, sum - root->val, curres, res);
             curres.pop_back();
         }
     }
 };

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