码迷,mamicode.com
首页 > 其他好文 > 详细

112. Path Sum (Tree; DFS)

时间:2015-10-04 11:02:20      阅读:115      评论:0      收藏:0      [点我收藏+]

标签:

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

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

              5
             /             4   8
           /   /           11  13  4
         /  \              7    2      1
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(!root) return false;
        flag = false;
        target = sum;
        preOrder(root,0);
        return flag;
        
    }
    void preOrder(TreeNode* node,int sum){
        sum = node->val + sum; //递归前,加上当前节点
        if(node->left)
        {           
            preOrder(node->left,sum);
        }
        if(node->right)
        {           
            preOrder(node->right,sum);
        }
        if(!node->left && !node->right && sum == target) //递归结束条件:到了叶子节点
        {
            flag = true;
        }
    }
private:
    bool flag;
    int target;
};

 

112. Path Sum (Tree; DFS)

标签:

原文地址:http://www.cnblogs.com/qionglouyuyu/p/4854200.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!