LeetCode 129 Sum Root to Leaf Numbers DFS

You are given the root of a binary tree containing digits from 0 to 9 only.

Each root-to-leaf path in the tree represents a number.

For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123.
Return the total sum of all root-to-leaf numbers. Test cases are generated so that the answer will fit in a 32-bit integer.

A leaf node is a node with no children.

Solution

还是利用 \(DFS\)：用另一个参数 \(cur\) 表示当前的值，对于 \(cur\) 的更新：

\[cur = 10\cdot cur+root.val \]

如果到达叶子节点，则直接返回 \(cur\). 最后对左右子树分别递归即可：

\[\text{dfs}(root.right,cur)+\text{dfs}(root.left,cur) \]

点击查看代码

/**
 * 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 {
private:
    
    int dfs(TreeNode* root, int ans){
        if(!root) return 0;
        ans = ans*10 + root->val;
        if(!root->left && !root->right)return ans;
        return dfs(root->left, ans)+dfs(root->right,ans);
    }
public:
    int sumNumbers(TreeNode* root) {
        if(root==NULL) return 0;
        return dfs(root, 0);
    }
};