Sum Root to Leaf Numbers

最新推荐文章于 2019-08-12 13:35:43 发布

原创最新推荐文章于 2019-08-12 13:35:43 发布 · 372 阅读

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本文介绍了一种算法，用于求解二叉树中所有从根节点到叶节点的路径数值之和。通过深度优先搜索（DFS）遍历二叉树，递归地计算路径值并累加到结果中。

/**
 * 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 getSum(TreeNode *t, int *res, int current) {
        current = current*10+t->val;
        if (t->left==NULL && t->right==NULL) {
            *res+=current;
            return;
        }
        if (t->left!=NULL) 
            getSum(t->left, res, current);
        if (t->right!=NULL) 
            getSum(t->right, res, current);
    }
    int sumNumbers(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root==NULL) {
            return 0;
        }
        int res=0;
        getSum(root, &res, 0);
        return res;
    }
};

The 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 dfs(TreeNode *root, int &res, int cur) {
        if (!root) return;
        int temp=cur*10+root->val;
        if (root->left==NULL && root->right==NULL) {
            res+=temp;
            return;
        }
        dfs(root->left, res, temp);
        dfs(root->right, res, temp);
        
        
    }
    int sumNumbers(TreeNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int res=0, cur=0;
        dfs(root, res, cur);
        return res;
    }
};