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Leetcode 437. Path Sum III 路径和3 解题报告

2016-10-23 14:29 399 查看

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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