113. Path Sum II && 面试题34. 二叉树中和为某一值的路径

Problem

113

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

Note: A leaf is a node with no children.

Example

Given the below binary tree and sum = 22,

return

面试题34

输入一棵二叉树和一个整数，打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。

Example

同上

Solution

/**
 * 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) {
        vector<vector<int>> ret;
        if(!root)
            return ret;
        vector<int> tmp;
        int curSum = 0;

        pathSumHelper(root,sum,curSum,tmp,ret);

        return ret;
        

    }

    void pathSumHelper(TreeNode* root,int sum,int &curSum,vector<int> &tmp，vector<vector<int>> &ret)
    {
        if(!root)
            return;
        
        curSum += root->val;
        tmp.push_back(root->val);

        if(!root->left && !root->right && curSum == sum)
        {
            ret.push_back(tmp);
        }

        if(root->left)
            pathSumHelper(root->left,sum,curSum,tmp,ret);
        if(root->right)
            pathSumHelper(root->right,sum,curSum,tmp,ret);

        curSum -= root->val;
        tmp.pop_back();
    }
};