Leetcode 112 Path Sum

triplebee

发布于 2018-01-12 14:53:19

4680

发布于 2018-01-12 14:53:19

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

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

判断二叉树是否有一条从根到叶的路径，使权值之和等于sum

/**
 * 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:
    bool dfs(TreeNode* root,int now,int sum)
    {
        now+=root->val;
        if(!root->left && !root->right) return now == sum;
        bool l=false,r=false;
        if(root->left) l=dfs(root->left,now,sum);
        if(root->right) r=dfs(root->right,now,sum);
        return l || r;
    }
    bool hasPathSum(TreeNode* root, int sum) {
        if(!root) return false;
        return dfs(root,0,sum);
    }
};

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原始发表：2016-10-12 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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