Given a binary tree containing digits from0-9only, each root-to-leaf path could represent a number. An example is the root-to-leaf path1->2->3which represents the number123. Find the total sum of all root-to-leaf numbers. Note: A leaf is a node with no children. Example: 1->212...
The root-to-leaf path1->3represents the number13. Return the sum = 12 + 13 =25. 代码: 1/**2* Definition for binary tree3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { val = x; }8* }9*/10publicclassSolution {11publicintsumN...
Find the total sum of all root-to-leaf numbers. For example, 1 / \ 2 3 The root-to-leaf path 1->2 represents the number 12. The root-to-leaf path 1->3 represents the number 13.Return the sum = 12 + 13 = 25...
FunctionhasRootToLeafSum (sum value S, root of Binary tree T)1. Set path to FALSE. Path represent the Boolean variable which determines whether there is a root to leaf pathor not. 2. Consider the base cases. • If (root==NULL&&S==0) //case a return true; • Subtract the ...
The root-to-leaf path 1->3 represents the number 13. Return the sum = 12 + 13 =25. 思路 求根节点到叶节点的路径组合数字之和,如上样例。 前序遍历该树,定义一个方法,表示当前节点到所有叶节点的组合数字和,其中当前值为sum。当节点为空时返回0,否则计算到此节点的组合数字;当该节点为叶节点时返...
the total sum of all root-to-leaf numbers */ public int sumNumbers(TreeNode root) { // write your code here List<Integer> result = new ArrayList<>(); int sum = 0; dfs(root, 0, result); for (int num : result) { sum += num; } return sum; } private void dfs(TreeNode root...
Find the total sum of all root-to-leaf numbers. Note:A leaf is a node with no children. Example: Input: [1,2,3] 1 / \ 2 3 Output: 25 Explanation: The root-to-leaf path 1->2 represents the number 12. The root-to-leaf path 1->3 represents the number 13. ...
public int sumNumbers(TreeNode root) { return getSum(root, 0); } private int getSum(TreeNode root, int curSum) { // condition if(root == null) { return 0; } curSum = curSum*10 + root.val; if(root.left==null && root.right==null) { return curSum; } // recursion return ...
The root-to-leaf path1->2represents the number12. The root-to-leaf path1->3represents the number13. Return the sum = 12 + 13 =25. 思路:本题可以按照先序遍历来求和,对于每个结点,采用 temp = value* 10 + node->val来更新值。注意先序遍历的写法,不要多考虑情况,一开始我加的判断太多了,...
=nil{result+=dfs(root.Left,pathSum+root.Left.Val)}// 如果右子结点不为空,则递归处理右子结点ifroot.Right!=nil{result+=dfs(root.Right,pathSum+root.Right.Val)}returnresult} 题目链接: Sum Root to Leaf Numbers : https://leetcode.com/problems/sum-root-to-leaf-numbers/...