LeetCode 113. Path Sum II

最新推荐文章于 2023-11-14 10:36:43 发布

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本文介绍了一种算法，该算法接收一个二叉树及一个整数目标和作为输入，然后找出所有从根节点到叶子节点且路径上节点值之和等于目标和的路径。通过递归遍历的方法实现，适用于数据结构与算法的学习。

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

/*
  Given a binary tree and a sum, find all root-to-leaf paths where each path's sum
  equals the given sum.
  For example:
  Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

  return

  [
     [5,4,11,2],
    [5,8,4,5]
  ]
*/
void pathSum(TreeNode* root, int sum, vector< vector<int> >& res, vector<int> path) { 
  if(!root) return;
  // push the node into path anyway....
  path.push_back(root->val); 
  if(!(root->right) && !(root->left) && (sum == root->val)) { 
    res.push_back(path); 
  } 
  pathSum(root->left, sum - root->val, res, path); 
  pathSum(root->right, sum - root->val, res, path); 
} 
 
vector< vector<int> > pathSum(TreeNode* root, int sum) { 
  if(!root) return {}; 
  vector< vector<int> > res; 
  vector<int> path; 
  pathSum(root, sum, res, path); 
  return res; 
}