leetcode:Path Sum (路径之和) 【面试算法题】
2013-10-17 11:15
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题目:
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
题意判断从根节点到叶子节点的路径和是否有等于sum的值。
叶子节点就是没有子节点的节点,判断叶子节点的当前值是否等于当前的sum。
若不是叶子节点,将当前sum减去当前节点的值,递归求解。
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->2which sum is 22.
题意判断从根节点到叶子节点的路径和是否有等于sum的值。
叶子节点就是没有子节点的节点,判断叶子节点的当前值是否等于当前的sum。
若不是叶子节点,将当前sum减去当前节点的值,递归求解。
/** * Definition for binary tree * 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)return false; if(sum==root->val&&root->left==NULL&&root->right==NULL)return true; return (root->left && hasPathSum(root->left,sum-root->val) ) ||(root->right && hasPathSum(root->right,sum-root->val)); } }; // blog.csdn.net/havenoidea
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