[leetcode] 113. Path Sum II

来源：互联网发布：arm7 java 编辑：程序博客网时间：2024/05/21 11:32

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]]

解法一：

/** * Definition for a binary tree node. * 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) {        if(!root) return {};        vector<vector<int>> res;        vector<int> out;        pathSum(res, out, root, sum);        return res;    }        void pathSum(vector<vector<int>>&res, vector<int>& out, TreeNode* root, int sum){        if(!root) return;                if(!root->left&&!root->right){            if(root->val==sum){                out.push_back(root->val);                res.push_back(out);                out.pop_back();                return;            }else return;        }                out.push_back(root->val);        pathSum(res, out, root->left, sum-root->val);        out.pop_back();                out.push_back(root->val);        pathSum(res, out, root->right, sum-root->val);        out.pop_back();    }};

解法二：

更加简洁点。

/** * Definition for a binary tree node. * 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) {        if(!root) return {};        vector<vector<int>> res;        vector<int> out;         out.push_back(root->val);        pathSum(res, out, root, sum);        return res;    }        void pathSum(vector<vector<int>>&res, vector<int>& out, TreeNode* root, int sum){        if(!root->left&&!root->right&&root->val==sum){            res.push_back(out);        }                if(root->left){            out.push_back(root->left->val);            pathSum(res, out, root->left, sum-root->val);            out.pop_back();        }                if(root->right){            out.push_back(root->right->val);            pathSum(res, out, root->right, sum-root->val);            out.pop_back();        }    }};

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