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…

https://leetcode.com/problems/sum-root-to-leaf-numbers/ 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…

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. For example, 1 / \ 2 3…

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. For example, 1 / \ 2 3…

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. For example, 1 / \ 2 3…

题目 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. For example, 1 / \…

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. For example, 1 / \ 2 3…

原题地址 二叉树的遍历 代码: vector<int> path; int sumNumbers(TreeNode *root) { if (!root) ; ; path.push_back(root->val); if (!root->left && !root->right) { ; i < path.size(); i++) sum = sum * + path[i]; } else { if (root->left) sum += sum…

树的数值为[0, 9], 每一条从根到叶子的路径都构成一个整数,(根的数字为首位),求所有构成的所有整数的和 深度优先搜索,通过一个参数累加整数 class Solution { public: void helper(TreeNode* node, int path, int& sum){ if(!node){ return; } //a(node) //lk("root",node) //a(path) //dsp if(!node->left &&…

DFS的标准形式 用一个String记录路径,最后判断到叶子时加到结果上. int res = 0; public int sumNumbers(TreeNode root) { if (root==null) return res; helper(root,""); return res; } public void helper(TreeNode root,String s) { s += root.val+""; if (root.left==null&…