題目描述

image.png

題解

解題思路與二叉樹(shù)根節(jié)點(diǎn)到葉節(jié)點(diǎn)的所有路徑和一題相似，都是采用遞歸算法。但這個(gè)題加了一點(diǎn)，要求保存路徑到vector中。

為了保存路徑，這里給遞歸函數(shù)傳遞一個(gè)vector類(lèi)型的參數(shù)，用于保存從根節(jié)點(diǎn)到當(dāng)前節(jié)點(diǎn)的路徑。將當(dāng)前節(jié)點(diǎn)值追加到vector末尾，再向下層傳遞。當(dāng)?shù)竭_(dá)葉節(jié)點(diǎn)時(shí)，判斷是否符合條件，如果符合，就將該路徑加到結(jié)果中。

到葉節(jié)點(diǎn)就可以完成判斷并得到路徑了，所以遞歸函數(shù)不需要返回值。

代碼

// pathSum.cpp

#include

#include

using namespace std;

struct TreeNode {

int val;

struct TreeNode *left;

struct TreeNode *right;

TreeNode(int val){

this->val = val;

this->left = NULL;

this->right = NULL;

}

};

class Solution {

public:

/**

*

* @param root TreeNode類(lèi)

* @param sum int整型

* @return int整型vector>

*/

vector > pathSum(TreeNode* root, int sum) {

// write code here

if (root == NULL){

return meet;

}

vector path;

targetSum = sum;

sumNumbers(root, root->val, path);

return meet;

}

private:

vector> meet;

int targetSum;

void sumNumbers(TreeNode* t, int sum, vector path){

path.push_back(t->val);

if (t->left == NULL && t->right == NULL){

if (sum == targetSum){

meet.push_back(path);

}

}

if (t->left != NULL){

sumNumbers(t->left, t->left->val + sum, path);

}

if (t->right != NULL){

sumNumbers(t->right, t->right->val + sum, path);

}

}

};

int main()

{

TreeNode* root = new TreeNode(5);

TreeNode* left = new TreeNode(4);

TreeNode* right = new TreeNode(8);

TreeNode* leftleft = new TreeNode(1);

TreeNode* leftright = new TreeNode(11);

TreeNode* leftrightleft = new TreeNode(2);

TreeNode* leftrightright = new TreeNode(7);

TreeNode* rightright = new TreeNode(9);

root->left = left;

root->right = right;

left->left = leftleft;

left->right = leftright;

leftright->left = leftrightleft;

leftright->right = leftrightright;

right->right = rightright;

Solution s;

vector> meet = s.pathSum(root, 22);

for (int i = 0; i < meet.size(); i++){

for (int j = 0; j < meet[i].size(); j++){

cout << meet[i][j] << " ";

}

cout << endl;

}

delete root;

delete left;

delete right;

delete leftleft;

delete leftright;

delete leftrightleft;

delete leftrightright;

delete rightright;

return 0;

}

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