LC 129.求根节点到叶节点数字之和
思路
pre_sum
代表之前改路径上所有数字的和dfs(root, pre_sum)
代表以root
为根节点的子树的所有数字之和dfs(当前节点) = (pre_sum * 10 + dfs(右子树)) + (pre_sum * 10 + dfs(左子树))
代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int sumNumbers(TreeNode* root) {
return dfs(root, 0);
}
int dfs(TreeNode* root, int pre_sum) {
if (!root) {
return 0;
}
// 如果是叶子节点
int ans = 0;
if (!root->left && !root->right) {
return pre_sum * 10 + root->val;
}
if (root->left) {
int cur_sum = pre_sum * 10 + root->val;
ans += dfs(root->left, cur_sum);
}
if (root->right) {
int cur_sum = pre_sum * 10 + root->val;
ans += dfs(root->right, cur_sum);
}
return ans;
}
};