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.
思路：
使用递归先序遍历。
代码如下：
/**
* 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(NULL == root)
return false;
return DFS(root,0,sum);
}
bool DFS(TreeNode * root,int curTotal,int sum)
{
if(NULL == root)
return false;
curTotal += root->val;
if( !root->left && !root->right && (curTotal == sum))
return true;
else
return DFS(root->left,curTotal,sum) || DFS(root->right,curTotal,sum);
}
};其他做法：
bool hasPathSum(TreeNode *root, int sum) {
if (root == NULL)
return false;
else if (root->left == NULL && root->right == NULL && root->val == sum)
return true;
else {
return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum - root->val);
}
}参考自：http://blog.csdn.net/booirror/article/details/42680111
2016-08-07 13:17:42
