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  • 112. Path Sum

    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.

    判断是否存在一条从根节点到叶子节点的路径,路径上节点之和为sum

    C++(6ms):

     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 == NULL) return false ;
    14         if (root->val == sum && root->left == NULL && root->right == NULL) return true ;
    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/mengchunchen/p/8047314.html
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