面试题34：二叉树中和为某一值的路径

考察二叉树的前序遍历。

C++版本

#include <iostream>
#include <vector>
#include <stack>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include "TreeNode.h"
using namespace std;

// vector<int>& path传递的是引用，会修改实参的值；也可以传递指针vector<int>* path
void FindPath(TreeNode* root, int expectedSum, vector<int>& path, int currentSum, vector<vector<int>>& ans){
    currentSum= currentSum + root->val;
    path.push_back(root->val);

    // 如果是叶子节点，并且路径上节点值的和等于输入的值，则添加这条路径
    bool isLeaf = root->left == nullptr && root->right == nullptr;
    if(isLeaf && currentSum == expectedSum){

        // 添加一行路径
        ans.push_back(vector<int>());
        cout<<"A path is found:";
        for(int i = 0; i < path.size(); i++){
            cout<<path[i]<<"\t";
            ans[ans.size() - 1].push_back(path[i]);
        }
        cout<<endl;
    }

    // 如果不是叶节点，则遍历它的子节点
    if(root->left != nullptr)
        FindPath(root->left, expectedSum, path, currentSum, ans);
    if(root->right != nullptr)
        FindPath(root->right, expectedSum, path, currentSum, ans);

    path.pop_back();
}

vector<vector<int>> FindPath(TreeNode* root,int expectNumber) {
    vector<vector<int>> ans;
    if(root == nullptr)
        return ans;
    vector<int> path;
    int currentSum = 0;
    FindPath(root, expectNumber, path, currentSum, ans);
    return ans;
}

int main()
{

    return 0;
}