Easy-题目36:112. Path Sum
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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 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.
题目大意:
给出一个二叉树和一个和值,判断是否存在一条从根节点到叶节点路径,使得路径上所有值之和等于给定的和值。
题目分析:
(1) 若该节点为空,则不存在;
(2) 若该节点为叶子节点,则判断节点值是否等于当前和值;
(3) 若不是叶子节点,则向左右子树分别递归搜索下去。
源码:(language:c)
bool hasPathSum(struct TreeNode* root, int sum) { if(!root) return 0; else if (!root->left && !root->right) return root->val==sum; else return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right,sum-root->val);}
成绩:
4ms,beats 73%,众数4ms,27%
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