博文中会简要介绍Leetcode P0129题目分析及解题思路。
“Sum Root to Leaf Numbers”这道题相对比较简单，本质上就是深度优先搜索，每次探索到叶子结点，然后将叶子结点组成的整数加到最终的和上。
Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3 which represents the number 123.
Find the total sum of all root-to-leaf numbers.
Note: A leaf is a node with no children.
以下是Java的题解代码实现。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int sumNumbers(TreeNode root) {
if (root == null)
return 0;
this.dfs(root, 0);
return this.sum;
}
private int sum = 0;
private void dfs(TreeNode curr, int prevSum) {
if (curr.left == null && curr.right == null) {
prevSum = prevSum*10+curr.val;
sum += prevSum;
}
if (curr.left != null)
this.dfs(curr.left, prevSum*10+curr.val);
if (curr.right != null)
this.dfs(curr.right, prevSum*10+curr.val);
}
}
以下是C++的题解代码实现。对于C++代码风格，后续将尽量遵循Google开源风格，包括命名规范等。
/**
* 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) {
if (!root)
return 0;
this->_Dfs(root, 0);
return this->sum_;
}
private:
int sum_ = 0;
void _Dfs(TreeNode *curr, int prev_sum) {
if (!curr->left && !curr->right) {
prev_sum = prev_sum*10+curr->val;
this->sum_ += prev_sum;
}
if (curr->left)
this->_Dfs(curr->left, prev_sum*10+curr->val);
if (curr->right)
this->_Dfs(curr->right, prev_sum*10+curr->val);
}
};