[LeetCode]Path Sum

最新推荐文章于 2020-07-07 13:07:47 发布

原创最新推荐文章于 2020-07-07 13:07:47 发布 · 744 阅读

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本文提供了一种使用深度优先搜索（DFS）解决树形结构中路径和问题的C++实现方法。详细阐述了如何通过递归遍历树节点，累积路径上的节点值，并检查是否达到目标和。

struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
	//if we need to find one path sum of a tree, then dfs is always a candidate
public:
	void DFS(TreeNode* root, int curSum, int sum, bool& OK)
	{
		if(!root || OK)
			return;
		curSum += root->val;
		if(!root->left && !root->right)
		{
			if(curSum == sum)
				OK = true;
			return;
		}
		DFS(root->left, curSum, sum, OK);
		DFS(root->right, curSum, sum, OK);
	}
    bool hasPathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
		bool ans = false;
        DFS(root, 0, sum, ans);
		return ans;
    }
};

second time

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void hasPathSumUtil(TreeNode* root, int sum, int curSum, bool& isExist)
    {
        if(isExist) return;
        if(root == NULL) return;
        if(root->left == NULL && root->right == NULL)
        {
           if(curSum+root->val == sum) isExist = true;
           return; 
        }
        hasPathSumUtil(root->left, sum, curSum+root->val, isExist);
        hasPathSumUtil(root->right, sum, curSum+root->val, isExist);
    }
    bool hasPathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        bool ans = false;
        hasPathSumUtil(root, sum, 0, ans);
        return ans;
    }
};