LeetCode|Path Sum*

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/DouMiaoO_Oo/article/details/51472865

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.

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if(root == NULL) return false;
        this->sum = sum;
        bool equal = false;
        helper(root, root->val, equal);
        return equal;
    }
private:
    int sum;
    void helper(TreeNode* root, int cur, bool& equal){
        if(equal) return;
        if(root->left == NULL && root->right == NULL) equal = (cur == sum);
        else {
            if(root->left != NULL) helper(root->left, cur + root->left->val, equal);
            if(root->right != NULL) helper(root->right, cur + root->right->val, equal);
        }        
    }
};

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]
]

class Solution {
public:
    vector<vector<int> > pathSum(TreeNode* root, int sum) {
        res.clear();
        if(root == NULL) return res;
        this->sum = sum;
        vector<int> v;
        helper(root, root->val, v);
        return res;
    }
private:
    vector<vector<int> > res;
    int sum;
    void helper(TreeNode* root, int cur, vector<int>& v){
        v.push_back(root->val);
        if(root->left == NULL && root->right == NULL){
            if(cur == sum) res.push_back(v);
        } else {
            if(root->left != NULL) helper(root->left, cur + root->left->val, v);
            if(root->right != NULL) helper(root->right, cur + root->right->val, v);
        } v.pop_back();
    }
};