* Path Sum III【任意起始点—终点路径和】

最新推荐文章于 2020-09-24 11:49:11 发布

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本文探讨了一道关于在二叉树中寻找路径之和等于特定值的问题，并提供了两种不同的解决方案。通过递归方法，文章详细解释了如何正确地统计所有符合条件的路径数量。

PROBLEM:

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

SOLVE:

/**
 * 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) {
        return pathSum1(root,sum,sum);
    }
private:
    int pathSum1(TreeNode* root, int sum, int total) {
        if(!root)
            return 0;
        else if(sum==root->val)
            return 1+pathSum2(root->left,0)+pathSum2(root->right,0)+pathSum1(root->left,total,total)+pathSum1(root->right,total,total);
        else
            return pathSum2(root->left,sum-root->val)+pathSum2(root->right,sum-root->val)+pathSum1(root->left,total,total)+pathSum1(root->right,total,total);
    }
    int pathSum2(TreeNode* root, int sum) {
        if(!root)
            return 0;
        else if(sum==root->val)
            return 1+pathSum2(root->left,0)+pathSum2(root->right,0);
        else
            return pathSum2(root->left,sum-root->val)+pathSum2(root->right,sum-root->val);
    }
};

注意：这题反反复复测试了好多次才做对......错误有：同一条轮径重复算入，或者漏掉了一些路径。

最终思路是：

1、创建两个版本的函数（Public函数是Leetcode的默认调用函数，其实可以去掉），pathSum1（）会分裂出一条新路径，而pathSum2（）只能在当前路径继续计算下去。（之前一种错误是只用了pathSum1（）函数，导致以第n层为起始点的路径被算了n-1次）

2、当sum与当前节点值相等时，除了返回 1 还要计算后续路径和为 0 的可能数。

大佬的答案：

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        if(!root) return 0;
        return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
    }
private:
    int sumUp(TreeNode* root, int pre, int& sum){
        if(!root) return 0;
        int current = pre + root->val;
        return (current == sum) + sumUp(root->left, current, sum) + sumUp(root->right, current, sum);
    }
};

大佬的程序确实简洁......