[Leetcode] Path Sum

最新推荐文章于 2020-07-07 13:07:47 发布

原创最新推荐文章于 2020-07-07 13:07:47 发布 · 128 阅读

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本文探讨了二叉树中寻找特定和的路径问题，包括确定是否存在从根到叶节点的路径使得路径上的数值之和等于给定值，以及找出所有这样的路径。通过递归算法实现解决方案。

Path Sum

AC Rate: 872/2770

My Submissions

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 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) {
        if (root == NULL) return false;
        if (root->left == NULL && root->right == NULL ) {
            if (root->val == sum) return true;
            else return false;
        }
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
    
    
};

Path Sum II

AC Rate: 691/2584

My Submissions

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]
]

/**
 * 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) {
        vector<vector<int> > res;
        if (root == NULL) return res;
        vector<int> v;
        gen(res, v, root, 0, sum);
        return res;
    }
    
    void gen(vector<vector<int> >& res, vector<int>& v, TreeNode* cur, int s, int sum) {
        v.push_back(cur->val);
        if (cur->left == NULL && cur->right == NULL) {
            if (s+cur->val == sum) res.push_back(v);
        }
        else {
            if (cur->left != NULL) gen(res, v, cur->left, s + cur->val, sum);
            if (cur->right!= NULL) gen(res, v, cur->right,s + cur->val, sum);
        }
        v.pop_back();
    }
};