LeetCode 112. Path Sum
2016-10-15 17:14
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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
return true, as there exist a root-to-leaf path
sum is 22.
题目的意思是说有没有一条从根结点到叶子结点的路径,使得路径上的数之和等于给定的数sum。
思路:
若当前结点为NULL,则返回false;
若当前结点不为NULL,进行如下判断:
当不是叶子结点的时候,应该递归为左结点+右节点,并且传递的参数sum需要减去val;
当为叶子结点的时候,需要进行如下判断:
当前值val是否和传递过来的sum是否相等?如果相等,则说明从根结点到该叶子结点路径上的所有val之和为sum,返回true;否则这条路径上的之和不为sum,返回false。
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->2which
sum is 22.
题目的意思是说有没有一条从根结点到叶子结点的路径,使得路径上的数之和等于给定的数sum。
思路:
若当前结点为NULL,则返回false;
若当前结点不为NULL,进行如下判断:
当不是叶子结点的时候,应该递归为左结点+右节点,并且传递的参数sum需要减去val;
当为叶子结点的时候,需要进行如下判断:
当前值val是否和传递过来的sum是否相等?如果相等,则说明从根结点到该叶子结点路径上的所有val之和为sum,返回true;否则这条路径上的之和不为sum,返回false。
** * 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 hasPathSum(TreeNode* root, int sum) { if(root==NULL){ return false; } else{ if(root->left==NULL && root->right==NULL){ if(sum==root->val) return true; else return false; } else{ bool flag1=false,flag2=false; if(root->left!=NULL) flag1 = hasPathSum(root->left,sum - root->val) ; if(root->right!=NULL) flag2 = hasPathSum(root->right,sum - root->val); return flag1+flag2; } } } };
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