面试100题004--二元树中找出和为某一值的所有路径

本文主要是介绍面试100题004--二元树中找出和为某一值的所有路径，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

一刷100题的第三天。

题目：输入一个整数和一棵二元树。
从树的根结点开始往下访问一直到叶结点所经过的所有结点形成一条路径。
打印出和与输入整数相等的所有路径。
例如输入整数22和如下二元树
10
/ /
5 12
/ /
4 7
则打印出两条路径：10, 12和10, 5, 7。

二元树节点的数据结构定义为：

struct BinaryTreeNode // a node in the binary tree
{
int m_nValue; // value of node
BinaryTreeNode *m_pLeft; // left child of node
BinaryTreeNode *m_pRight; // right child of node
};

答案：

#include <iostream>
#include <vector>
using namespace std;
struct BSTreeNode {
int m_nvalue;
BSTreeNode *m_pleft;
BSTreeNode *m_pright;
};
void addBSTreeNode(BSTreeNode * & pCurrent, int value) {
if (pCurrent == NULL) {
BSTreeNode *pBSTree = new BSTreeNode();
pBSTree->m_nvalue = value;
pBSTree->m_pleft = NULL;
pBSTree->m_pright = NULL;
pCurrent = pBSTree;
}
else {
if (pCurrent->m_nvalue > value) {
addBSTreeNode(pCurrent->m_pleft, value);
}
else if (pCurrent->m_nvalue < value) {
addBSTreeNode(pCurrent->m_pright, value);
}
else {
cout << "this tree has this number!" << endl;
}
}
}
void findPath(BSTreeNode * & pCurrent, int expectedSum, std::vector<int> &path,int &currentSum) {
if (!pCurrent) {
return;
}
currentSum += pCurrent->m_nvalue;
path.push_back(pCurrent->m_nvalue);
bool isLeaf = (!pCurrent->m_pleft&&!pCurrent->m_pright);
if (currentSum == expectedSum&&isLeaf) {
std::vector<int>::iterator iter = path.begin();
for (; iter != path.end();iter++) {
std::cout << *iter << '\t';
}
std::cout << std::endl;
}
if (pCurrent->m_pleft) {
findPath(pCurrent->m_pleft, expectedSum, path,currentSum);
}
if (pCurrent->m_pright) {
findPath(pCurrent->m_pright, expectedSum, path,currentSum);
}
currentSum -= pCurrent->m_nvalue;
path.pop_back();
}
int main() {
BSTreeNode *pRoot = NULL;
addBSTreeNode(pRoot, 10);
addBSTreeNode(pRoot, 5);
addBSTreeNode(pRoot, 4);
addBSTreeNode(pRoot, 7);
addBSTreeNode(pRoot, 12);
std::vector<int> path;
int currentSum = 0;
int expectedSum = 19;
findPath(pRoot, expectedSum, path, currentSum);
return 0;
}