112. Path Sum
problem description
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.
Note: A leaf is a node with no children.
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.
algorithm thought
直接递归解决,每次进入子节点的时候,将sum值减去当前节点的值,最后判断如果是叶子节点并且当前值为0,就返回true,否则返回false
code
/**
* 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;
sum-=root->val;
if(root->left==NULL&&root->right==NULL){
if(sum)
return false;
return true;
}
return hasPathSum(root->left,sum)||hasPathSum(root->right,sum);
}
};
algorithm analysis
遍历问题,时间复杂度O(n)
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