Leetcode 437. Path Sum III 路径和3 解题报告
2016-10-23 14:29
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1 解题思想
这道题就是给了一个二叉树和一个目标和sum找出所有路径,这个路径的和等于sum,只允许从父节点到子节点的路线
所以方法么,也就是最基本的dfs,不多说,对了可以看看之前相关的题目:
Leetcode 64. Minimum Path Sum 最小路径和 解题报告
Leetcode 112. Path Sum 路径和 解题报告
Leetcode 113. Path Sum II 路径和2 解题报告
2 原题
You are given a binary tree in which each node contains an integer value. Find the number of paths that sum to a given value. The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes). The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000. Example: root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10 / \ 5 -3 / \ \ 3 2 11 / \ \ 3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3 2. 5 -> 2 -> 1 3. -3 -> 11
3 AC解
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public int pathSum(TreeNode root, int sum) { if(root == null) return 0; return dfs(root, sum)+pathSum(root.left, sum)+pathSum(root.right, sum); } private int dfs(TreeNode root, int sum){ int res = 0; if(root == null) return res; if(sum == root.val) res++; res+=dfs(root.left,sum - root.val); res+=dfs(root.right,sum - root.val); return res; } }
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