leetcode 119: Path Sum II

最新推荐文章于 2021-01-15 11:49:21 发布

原创最新推荐文章于 2021-01-15 11:49:21 发布 · 658 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

216 篇文章

订阅专栏

本文介绍了一种算法，用于查找二叉树中所有从根节点到叶子节点的路径，其中路径上节点值的总和等于指定的目标值。通过递归方法实现，详细展示了如何遍历二叉树并收集符合条件的路径。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Path Sum IIOct 14 '12838 / 2481

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]
]

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        vector<vector<int> > res;
        vector<int> temp;
        if(root!=nullptr) 
            sumRec(res, temp, root, 0, sum);
        else
            return res;
    }
    
    void sumRec(vector<vector<int> > & res, vector<int>&  temp ,TreeNode* root, int cur, const int sum) {
        cur += root->val;
        temp.push_back(root->val);
        if(root->left==nullptr && root->right==nullptr) {
            if(cur==sum) {
                res.push_back(temp);
            } 
        }
        
        if(root->left != nullptr) {
            sumRec(res, temp, root->left, cur, sum);
        } 
        
        if(root->right!=nullptr) {
            sumRec(res,temp,root->right, cur, sum);
        } 
        temp.pop_back();
        cur-=root->val;
    }
};

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        //dfs   preorder
        vector<vector<int>> res;
        vector<int> temp;
        if(root==nullptr) return res;
        sumRec(res, temp, root, sum, 0);
        return res;
    }
    
    void sumRec(vector<vector<int>>& res, vector<int>& temp, TreeNode * root, const int sum, int cur) {
        cur += root->val;
        temp.push_back(root->val);
            
        if(root->left==nullptr && root->right==nullptr) {
            if(sum==cur) {
                res.push_back( temp);
            }
            return;
        }
        
        if(root->left!=nullptr) {
            sumRec(res, temp, root->left, sum, cur);
            temp.pop_back();
        }
        if(root->right!=nullptr) {
            sumRec(res, temp, root->right, sum, cur);
            temp.pop_back();
        }
    }
};