Leetcode 113. Path Sum II

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题目描述

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]]

输入：树，目标和
输出：从根到叶满足目标和的路径

思路

深度有限搜索，同时维护一个根到当前节点路径，并且在叶子节点处进行判断；检查是否符合目标和，如果符合则将路径添加到输出结果中。此处，如果能确定树中节点值均为正数的话，可以通过剪枝操作减少运算量。

代码

/** * 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> > res;    vector<vector<int> > pathSum(TreeNode* root, int sum) {        if(root==NULL) return res;        vector<int> emp;        dfs(root,emp,sum);        return res;    }    void dfs(TreeNode* root, vector<int> t, int sum){        if(root->left==NULL&&root->right==NULL){            if(root->val==sum){                t.push_back(root->val);                res.push_back(t);                return;            }        }else{            t.push_back(root->val);            if(root->left!=NULL){                dfs(root->left,t,sum-root->val);            }            if(root->right!=NULL){                dfs(root->right,t,sum-root->val);            }        }    }};