LEETCODE--Path Sum III

最新推荐文章于 2021-12-24 23:44:26 发布

prettysky123

最新推荐文章于 2021-12-24 23:44:26 发布

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本文介绍了一种算法，用于查找给定二叉树中所有路径之和等于特定值的数量。路径可以从任意节点开始，向下经过子节点，直至叶子节点。文章详细解释了递归算法实现，并提供了一个具体的例子来说明如何应用该算法。

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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:

5 -> 3
5 -> 2 -> 1
-3 -> 11

/**
 * 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 {
private:
    int Thispoint(TreeNode* root , int valsum ,int sum){
        if( !root)
            return 0;
        valsum += root->val;
        return (valsum == sum) + Thispoint(root->right, valsum, sum) + Thispoint(root->left, valsum, sum);
    }
public:
    int pathSum(TreeNode* root, int sum) {
        if( !root)
            return 0;
        return Thispoint (root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
    }
};