【leetcode】Path Sum I & II（middle）

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.

 

思路：树的题目，整体思路就是递归查找。三行解决。

bool hasPathSum(TreeNode *root, int sum) {

        if(root == NULL) return false;

        if(sum == root->val && (root->left == NULL) && (root->right == NULL)) return true; //和相等 且 是叶子结点



        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);

    }

 

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]

]

 

思路：找所有路径，也是用递归。发现满足的路径就压入。 

vector<vector<int> > pathSum(TreeNode *root, int sum) {

        vector<vector<int> > ans;

        vector<int> tmpans;

        findSum(root, sum, tmpans, ans);

        return ans;



    }



    void findSum(TreeNode *root, int sum, vector<int> tmpans, vector<vector<int>> &ans)

    {

        if(root == NULL) return;

        if(sum == root->val && (root->left == NULL) && (root->right == NULL)) //满足条件 压入答案

        {

            tmpans.push_back(root->val);

            ans.push_back(tmpans);

        }

        else

        {

            tmpans.push_back(root->val);

        }

        findSum(root->left, sum - root->val, tmpans, ans);

        findSum(root->right, sum - root->val, tmpans, ans);

    }