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.
解题思路:
由于需要找寻根到叶子节点的路径和,我们可以通过遍历一颗树即可得知,通常树的遍历有
四种方式:先序遍历、中序遍历、后序遍历、层次遍历,任选一种遍历方式即可.
解题代码(非递归):
/**
* Definition for binary tree
* 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)
return false;
queue<pair<TreeNode *,long long> > que;
que.push(make_pair(root,root->val));
while(!que.empty())
{
pair<TreeNode *,long long> p = que.front();
que.pop();
if(p.first->left == NULL && p.first->right == NULL && p.second == sum)
return true;
if(p.first->left)
que.push(make_pair(p.first->left,p.second + p.first->left->val));
if(p.first->right)
que.push(make_pair(p.first->right,p.second + p.first->right->val));
}
return false;
}
};
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool dfs(TreeNode *rt,long long sum)
{
if (rt->left == rt->right && !rt->left)
return sum == rt->val ;
if (rt->left && rt->right)
return dfs(rt->left,sum - rt->val) | dfs(rt->right,sum - rt->val);
return dfs(rt->left ? rt->left : rt->right, sum - rt->val);
}
bool hasPathSum(TreeNode *root, int sum)
{
return root ? dfs(root,sum) : false ;
}
};
LeetCode:Path Sum,布布扣,bubuko.com
原文:http://blog.csdn.net/dream_you_to_life/article/details/30339331