• Leetcode 112. Path Sum


    112. Path Sum

    • Total Accepted: 112241
    • Total Submissions: 352808
    • Difficulty: Easy

    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.

    思路:DFS。只要找到叶节点的时候,sum==0,就可以返回true;否则返回false。

    代码:

     1 /**
     2  * Definition for a binary tree node.
     3  * struct TreeNode {
     4  *     int val;
     5  *     TreeNode *left;
     6  *     TreeNode *right;
     7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     8  * };
     9  */
    10 class Solution {
    11 public:
    12     bool hasPathSum(TreeNode* root, int sum) {
    13         if(!root) return false;
    14         if(!root->left&&!root->right) return root->val==sum;//判断是否为叶节点
    15         return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);
    16     }
    17 };
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  • 原文地址:https://www.cnblogs.com/Deribs4/p/5686922.html
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