LeetCode 113: Path Sum II

最新推荐文章于 2024-05-22 20:39:12 发布

claviclelau

最新推荐文章于 2024-05-22 20:39:12 发布

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本文探讨了如何确定一棵二叉树中是否存在从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定的目标值。通过深度优先搜索(DFS)递归算法实现解决方案。

Difficulty: 2

Frequency: 2

Problem:

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and

sum
 = 22

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Solution:

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> > v_result;
        vector<int> v_current;
        DFS(root, sum, v_current, v_result);
        return v_result;
    }
    void DFS(TreeNode * root, int sum, vector<int> & v_current, vector<vector<int>> & v_result)
    {
        if (root==NULL)
            return;
        if (root->left==NULL&&root->right==NULL)
        {
            if (root->val==sum)
            {
                v_current.push_back(sum);
                v_result.push_back(v_current);
                v_current.pop_back();
            }
            return;
        }
        
        v_current.push_back(root->val);
        
        if (root->left!=NULL)
        {
            DFS(root->left, sum-root->val, v_current, v_result);
        }
        
        if (root->right!=NULL)
        {
            DFS(root->right, sum-root->val, v_current, v_result);
        }
        v_current.pop_back();
    }
};