112. Path Sum
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:
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/**
* 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);
}
};
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其他做法:
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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
本文转自313119992 51CTO博客,原文链接:http://blog.51cto.com/qiaopeng688/1835329