编程思路总结——递归

1. 二叉树中和为某一值的路径

路径：从树的根节点到叶子节点经过的节点形成的路径，例如途中（10，5，4），（10，5，7），（10，12）

满足和为22的路径有（10，5，7）、（10，12）

参考代码

void FindPath(TreeNode *root, vector<TreeNode> &vec, int cur, int aim)
{
    if (root == NULL)
        return;
    vec.push_back(*root);
    cur += root->val;
    bool IsLeft = ((root->left == NULL) && (root->right == NULL));
    if (cur == aim && IsLeft)
    {
        for (vector<TreeNode>::iterator beg = vec.begin(); beg != vec.end(); ++beg)
            cout << beg->val << " ";
        cout << endl;
    }
    if (root->left != NULL)
        FindPath(root->left, vec, cur, aim);
    if (root->right != NULL)
        FindPath(root->right, vec, cur, aim);
    vec.pop_back();
}
void FindPath(TreeNode *root, int aim)
{
    if (root == NULL)
        return;
    vector<TreeNode> vec;
    FindPath(root, vec, 0, aim);
}

测试

#include <iostream>
#include <vector>
using namespace std;
struct TreeNode 
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int v) : val(v), left(NULL), right(NULL) {};
};

void FindPath(TreeNode *root, vector<TreeNode> &vec, int cur, int aim)
{
    if (root == NULL)
        return;
    vec.push_back(*root);
    cur += root->val;
    bool IsLeft = ((root->left == NULL) && (root->right == NULL));
    if (cur == aim && IsLeft)
    {
        for (vector<TreeNode>::iterator beg = vec.begin(); beg != vec.end(); ++beg)
            cout << beg->val << " ";
        cout << endl;
    }
    if (root->left != NULL)
        FindPath(root->left, vec, cur, aim);
    if (root->right != NULL)
        FindPath(root->right, vec, cur, aim);
    vec.pop_back();
}
void FindPath(TreeNode *root, int aim)
{
    if (root == NULL)
        return;
    vector<TreeNode> vec;
    FindPath(root, vec, 0, aim);
}

int main()
{
    TreeNode *root = new TreeNode(10);
    TreeNode *p1 = new TreeNode(5);
    TreeNode *p2 = new TreeNode(12);
    TreeNode *p3 = new TreeNode(4);
    TreeNode *p4 = new TreeNode(7);
    root->left = p1;
    root->right = p2;
    p1->left = p3;
    p1->right = p4;
    FindPath(root, 22);
}