起初是为了方便使用递归,然后递归参数使用当前节点的左右子节点,和sum值减去该节点的值。具体如下: 1/**2* Definition for a binary tree node.3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10cl...
【LeetCode】Path Sum ---LeetCode java 小结 Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example: Given the below binary tree andsum = 22, 5 / \ 4 8 / / \ ...
bool dfs(TreeNode *node, int sum, int curSum) { if (node == NULL) return false; if (node->left == NULL && node->right == NULL) return curSum + node->val == sum; return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val); } bo...
Left, remainSum) { return true } // 当右子结点存在,且存在一条右子结点到叶子路径上所有值到和为 remainSum ,则满足题意 if root.Right != nil && hasPathSum(root.Right, remainSum) { return true } // 左右子结点都不满足题意,返回 false return false } 题目链接: Path Sum: leetcode.com...
437. Path Sum III 题目描述 给你一棵二叉树和一个整数sum,判断这棵树是否存在根节点到叶子节点,以至于路径上面的数和为sum。 例子 给定下面的二叉树和sum = 22, 返回true,因为存在根节点到叶子节点路径5->4->11->2,和为22。 解题思路 这个题目就是典型利用递归就可以解决,从根节点往下进行递归。
题目链接:Path Sum II | LeetCode OJ Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22, 5 /\ 48 //\ 11134 /\/\ ...
题目地址:https://leetcode.com/problems/path-sum/#/description 题目描述 Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. ...
if(sum==0) count++; path(root.left,sum); path(root.right,sum); } } 113. 路径总和 II 给定一个二叉树和一个目标和,找到所有从根节点到叶子节点路径总和等于给定目标和的路径。 说明:叶子节点是指没有子节点的节点。 示例: 给定如下二叉树,以及目标和sum = 22, ...
int pathSum = 0; TreeNode prev = null; TreeNode curr = root; while (curr != null || !s.isEmpty()){ // go down all the way to the left leaf node // add all the left nodes to the stack while (curr != null){ s.push(curr); ...
intminPathSum(vector<vector<int>>&grid){ minSum=(~(unsignedint))>>; my_grid=grid; rowMax=grid.size(); colMax=grid[].size(); tra(,,); returnminSum; } }; 解法2:DP(还是不熟练,不太熟练递推dp和递归dp的区别,参考文章) dp[100][100];该dp数组记录的是每个位置上的最优解,即到达这一...