leetcode 113 Path Sum II

最新推荐文章于 2020-06-08 14:28:23 发布

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本文介绍了一种算法，用于在给定二叉树和指定和的情况下找到所有从根节点到叶子节点的路径，使得每条路径上的节点值之和等于给定的值。通过递归方法实现，并附带示例代码。

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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22 ,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

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/**
 * 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:
   void findPath(vector<vector<int>> &res, vector<int> vec, TreeNode *root, int remain) {
		if(root!=NULL && root->left==NULL && root->right==NULL) {
			if(remain == root->val) {
				vec.push_back(root->val);
				res.push_back(vec);
			}
			return;
		}

		if(root==NULL) return ;
		//if(root!=NULL && root->val > remain) return;
        vec.push_back(root->val);
		findPath(res, vec, root->left, remain-root->val);
		findPath(res, vec, root->right, remain-root->val);
		
		return;
	}
	vector<vector<int>> pathSum(TreeNode *root, int sum) {
		vector<vector<int>> res;
		vector<int> vec;

		findPath(res, vec, root, sum);

		return res;
	}
};