LeetCode 437 Path Sum III
2017-08-05 18:52
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题目:
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
题目链接
题意:
给一棵二叉树,其中每个节点包含一个整数值。给定一个值sum,要求找到路径和为sum的路径的个数。
路径不需要从root开始或结束,但必须向下(只能从父节点到子节点)
数的节点数不超过1000个,取值范围在-1000000到1000000之间。
由于路径并不要求必须以叶子为终点,所以不能写从叶子向上递加的算法,通过观察得出,每一个非叶子结点都可以作为一段路径的起点,而每一个非根节点结点都可以作为一段路经的终点,所以我们可以遍历整个树,枚举所有可能的起点,对与每一个起点进行搜索,以其开始的路径是否有长度为sum的,有的话,ans++,当所有遍历结束,返回ans。
代码如下:
class Solution {
public:
int target, ans = 0;
void upToSum(TreeNode* node, int sum) {
if (!node) return;
sum += node->val;
if (sum == target) {
ans++;
}
upToSum(node->left, sum);
upToSum(node->right, sum);
}
void dfs(TreeNode * node) {
if (!node) return;
upToSum(node, 0);
dfs(node->left);
dfs(node->right);
}
int pathSum(TreeNode* root, int sum) {
target = sum;
dfs(root);
return ans;
}
};
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10 / \ 5 -3 / \ \ 3 2 11 / \ \ 3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3 2. 5 -> 2 -> 1 3. -3 -> 11
题目链接
题意:
给一棵二叉树,其中每个节点包含一个整数值。给定一个值sum,要求找到路径和为sum的路径的个数。
路径不需要从root开始或结束,但必须向下(只能从父节点到子节点)
数的节点数不超过1000个,取值范围在-1000000到1000000之间。
由于路径并不要求必须以叶子为终点,所以不能写从叶子向上递加的算法,通过观察得出,每一个非叶子结点都可以作为一段路径的起点,而每一个非根节点结点都可以作为一段路经的终点,所以我们可以遍历整个树,枚举所有可能的起点,对与每一个起点进行搜索,以其开始的路径是否有长度为sum的,有的话,ans++,当所有遍历结束,返回ans。
代码如下:
class Solution {
public:
int target, ans = 0;
void upToSum(TreeNode* node, int sum) {
if (!node) return;
sum += node->val;
if (sum == target) {
ans++;
}
upToSum(node->left, sum);
upToSum(node->right, sum);
}
void dfs(TreeNode * node) {
if (!node) return;
upToSum(node, 0);
dfs(node->left);
dfs(node->right);
}
int pathSum(TreeNode* root, int sum) {
target = sum;
dfs(root);
return ans;
}
};
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