LeetCode: Path Sum II [113]

时间：2014-06-20 10:53:08 收藏：0 阅读：181

【题目】

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]
]

【题意】

判断二叉树中是否存在一条从根到叶子节点的路径，使得路径上的节点值之和等于所给的值，输出所有的路径

【思路】

DFS，递归

【代码】

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:

    void dfs(TreeNode*root, vector<vector<int> >&result, vector<int>path, int pathSum, int sum){
        path.push_back(root->val);
        pathSum+=root->val;
        if(root->left==NULL && root->right==NULL){
            if(pathSum==sum) result.push_back(path);
            return;
        }
        
        //搜索左子树
        if(root->left)
            dfs(root->left, result, path, pathSum, sum);
        //搜索右子树
        if(root->right)
            dfs(root->right, result, path, pathSum, sum);
    }

    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        vector<vector<int> >result;
        if(root==NULL)return result;
        
        vector<int>path;
        int pathSum=0;
        dfs(root, result, path, pathSum, sum);
        return result;
    }
};

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