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leetcode/algorithms/cpp/pathSum/pathSum.cpp at master · wcoder0/leetcode · GitHub
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//
Source : https://oj.leetcode.com/problems/path-sum/
//
Author : Hao Chen
//
Date : 2014-06-22
/*
*********************************************************************************
*
* 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.
*
*********************************************************************************
*/
#
include
<
time.h
>
/*
*
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class
Solution
{
public:
Solution
(){
srand
(
time
(
NULL
));
}
bool
hasPathSum
(TreeNode *root,
int
sum) {
return
hasPathSum3
(root, sum,
0
);
return
hasPathSum2
(root, sum);
return
hasPathSum1
(root, sum);
}
bool
hasPathSum3
(TreeNode* root,
int
sum,
int
s) {
if
( root ==
NULL
)
return
false
;
s += root->
val
;
if
( !root->
left
&& !root->
right
)
return
s == sum;
return
(
hasPathSum3
(root->
left
, sum, s) ||
hasPathSum3
(root->
right
, sum, s));
}
bool
hasPathSum1
(TreeNode *root,
int
sum) {
if
(root==
NULL
)
return
false
;
vector<TreeNode*> v;
v.
push_back
(root);
while
(v.
size
()>
0
){
TreeNode* node = v.
back
();
v.
pop_back
();
if
(node->
left
==
NULL
&& node->
right
==
NULL
){
if
(node->
val
== sum){
return
true
;
}
}
if
(node->
left
){
node->
left
->
val
+= node->
val
;
v.
push_back
(node->
left
);
}
if
(node->
right
){
node->
right
->
val
+= node->
val
;
v.
push_back
(node->
right
);
}
}
return
false
;
}
bool
hasPathSum2
(TreeNode *root,
int
sum) {
if
(root==
NULL
)
return
false
;
if
(root->
left
==
NULL
&& root->
right
==
NULL
){
return
(root->
val
==sum);
}
if
(root->
left
){
root->
left
->
val
+= root->
val
;
if
(
hasPathSum2
(root->
left
, sum)){
return
true
;
}
}
if
(root->
right
){
root->
right
->
val
+= root->
val
;
if
(
hasPathSum2
(root->
right
, sum)){
return
true
;
}
}
return
false
;
}
};
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