LeetCode:437. Path Sum III

Posted tacia

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了LeetCode:437. Path Sum III相关的知识,希望对你有一定的参考价值。

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

给定一颗二叉树,找出路径和等于sum的路径数。这里的路径是指从父节点到子节点的路径,不必一定是从根节点到叶子节点。马上想到深度优先搜索,但是要注意路径不一定从根节点开始。所以对于一个父节点来说,它所在路径上满足题设要求的路径,除了包括左右子树中路径和为sum-root->val的路径外,还应包括左右子树中路径和为sum的路径。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        if(!root)
          return 0;
        return pathSum(root->left,sum)+pathSum(root->right,sum)+pathSumRoot(root,sum);
    }
    
    int pathSumRoot(TreeNode* root, int sum)
    {
        if(!root)
          return 0;
        int count=(sum==root->val?1:0);
        return pathSumRoot(root->left,sum-root->val)+ pathSumRoot(root->right,sum-root->val)+count;
    }
};

 

以上是关于LeetCode:437. Path Sum III的主要内容,如果未能解决你的问题,请参考以下文章

leetcode 437. Path Sum III

LeetCode - 437. Path sum III

leetcode 437. Path Sum III

leetcode 437 路径和 Path Sum III

[LeetCode] 437. Path Sum III_ Easy tag: DFS

LeetCode 437. Path Sum III(统计路径和等于sum的路径数量)