leetcode 437. Path Sum III
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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
解法1,DFS:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: int
# java solution
public class Solution {
public int pathSum(TreeNode root, int sum) {
if (root == null) return 0;
return pathSumFrom(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
}
private int pathSumFrom(TreeNode node, int sum) {
if (node == null) return 0;
return (node.val == sum ? 1 : 0)
+ pathSumFrom(node.left, sum - node.val) + pathSumFrom(node.right, sum - node.val);
}
}
"""
def dfs(node, target):
if not node: return 0
return int(node.val == target)+dfs(node.left, target-node.val)+dfs(node.right, target-node.val)
if not root: return 0
return dfs(root, sum)+self.pathSum(root.left, sum)+self.pathSum(root.right, sum)
尤其注意是dfs(root, sum)+self.pathSum(root.left, sum)+self.pathSum(root.right, sum) 而不是dfs+dfs+dfs!容易出错!
另外解法就是前缀树的解法,前缀树记录sum,比如:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1
到节点3的前缀sum就是{10:1, 15:1, 18:1},则包括节点3的满足sum=8的路径就是{10,5,3}这条通路减去{10}这条通路,两个前缀相减,更深的前缀-浅的前缀就是中间路路径,18-10=8满足条件。
代码:
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: int
"""
def traverse(node, total, pre_sum):
if not node:
return 0
total += node.val
ans = pre_sum.get(total-sum, 0)
pre_sum[total] = pre_sum.get(total, 0)+1
ans += traverse(node.left, total, pre_sum) + traverse(node.right, total, pre_sum)
pre_sum[total] -= 1
return ans
return traverse(root, 0, {0:1})
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