• Path Sum II


    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]]

    分析: dfs求解即可

    /**
     * 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:
        void find(TreeNode* root, int sum,vector<int>& curPath, vector<vector<int>> & res ){
            if(root==nullptr)
                return;
            if(root->left){
                curPath.push_back(root->left->val);
                find(root->left, sum-root->val, curPath,res);
                curPath.pop_back();
            }
            if(root->right){
                curPath.push_back(root->right->val);
                find(root->right, sum-root->val,curPath,res);
                curPath.pop_back();
            }
            if(root->left==nullptr && root->right==nullptr && sum==root->val)
                res.push_back(curPath);
            
            return;
        }
        vector<vector<int>> pathSum(TreeNode* root, int sum) {
            vector<vector<int>> res;
            if(root==nullptr)
                return res;
            vector<int> curPath;
            curPath.push_back(root->val);
            find(root, sum, curPath, res);
            
            return res;
        }
    };
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  • 原文地址:https://www.cnblogs.com/willwu/p/6057437.html
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