剑指Offer算法实现之二十五：二叉树中和为某一值的路径

题目：输入二叉树和一个整数，打印出二叉树中节点值的和为输入整数的所有路径。从数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。二叉树的接待你定义如下：

struct BinaryTreeNode {
    int m_nValue;
    BinaryTreeNode *m_pLeft;
    BinaryTreeNode *m_pRight;
};

思路：

递归，函数参数中维护”路径“，回溯。

编译环境：ArchLinux+Clang3.3， C++11

实现一：

#include <iostream>
#include <vector>
#include <cstring>
#include <cctype>
using namespace std;

struct BinaryTreeNode {
    int m_nValue;
    BinaryTreeNode *m_pLeft;
    BinaryTreeNode *m_pRight;
};
void do_printPath(BinaryTreeNode *pRoot, vector<int>& path, int expectedSum, int pathSum);
/** Wrapper: 打印符合条件的路径 *//
void printPath(BinaryTreeNode *pRoot, int expectedSum)
{
    if (pRoot == nullptr) return;
    vector<int> path;
    do_printPath(pRoot, path, expectedSum, 0);
}
void do_printPath(BinaryTreeNode *pRoot, vector<int>& path, int expectedSum, int pathSum)
{
    path.push_back(pRoot->m_nValue);
    pathSum += pRoot->m_nValue;
    
    bool isLeaf = !pRoot->m_pLeft && !pRoot->m_pRight;
    if (isLeaf && pathSum == expectedSum) {
        for (auto i : path) {
            cout << i << ' ';
        }
        cout << endl;
    }
/** 递归 **/
    if (pRoot->m_pLeft) 
        do_printPath(pRoot->m_pLeft, path, expectedSum, pathSum);
    if (pRoot->m_pRight)
        do_printPath(pRoot->m_pRight, path, expectedSum, pathSum);
    path.pop_back(); // 回溯

}

int getNum(const char *start, const char *end, int *moved)
{
    int n = 0;
    while (start < end && isdigit(*start)) {
        (*moved)++;
        n *= 10;
        n += *start - '0';
        start++;
    }
    return n;
}

/** 
 * 根据二叉树的字符表示形式(形如"1(2,3(,4))")创建二叉树 
 * 区间：[start, end)
**/
BinaryTreeNode *createTree(const char *start, const char *end)
{
    if (start >= end) {
        return nullptr;
    }
//    if (end - start == 1) {
//        return new BinaryTreeNode{*start-'0', nullptr, nullptr};
//    }
    int rootLen = 0;
    int rootValue = getNum(start, end, &rootLen);
    if (start + rootLen == end) {
        return new BinaryTreeNode{rootValue, nullptr, nullptr};
    }
    const char *start1 = start + rootLen + 1;
    const char *end1 = start + rootLen + 1;
    int cnt = 0;
    while (true) {
        if (*end1 == '(') cnt++;
        if (*end1 == ')') cnt--;
        if (*end1 == ',' && cnt == 0) break;
        end1++;
    }
    const char *start2 = end1+1;
    const char *end2 = end1+1;
    cnt = 0;
    while (true) {
        if (*end2 == '(') cnt++;
        if (*end2 == ')' && cnt == 0) break;
        if (*end2 == ')') cnt--;
        end2++;
    }
    return new BinaryTreeNode{rootValue, 
                    createTree(start1, end1),
                    createTree(start2, end2)};
}
BinaryTreeNode *createTree(const char *str)
{
    return  createTree(str, str + strlen(str));
}
/** 打印二叉树 **/
void printTree(BinaryTreeNode *pRoot){
    if (!pRoot) return;
    cout << pRoot->m_nValue;
    if (pRoot->m_pLeft || pRoot->m_pRight) cout << '(';
    if (pRoot->m_pLeft) {
        printTree(pRoot->m_pLeft);
    }
    if (pRoot->m_pLeft || pRoot->m_pRight) cout << ',';
    if (pRoot->m_pRight) {
        printTree(pRoot->m_pRight);
    }
    if (pRoot->m_pLeft || pRoot->m_pRight) cout << ')';
}

int main()
{
    BinaryTreeNode *pRoot = createTree("10(5(4,7),12)");
    printTree(pRoot); cout << endl;
    printPath(pRoot, 22);
}