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# 437.Path-Sum-III

## 题目描述

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

## 代码

### Approach 1: DFS

path sum = 经过root节点的sum path + 左边root.left节点的sum path +右边root.right节点sum path
Time Complexity should be O(N^2) for the worst case and O(NlogN) for balanced binary Tree.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int pathSum(TreeNode root, int sum) {
if (root == null) return 0;
// 中 + 左 + 右
return dfs(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
}
// 包含当前 node 的 path sum
private int dfs(TreeNode node, int sum) {
if (node == null) return 0;
return (node.val == sum ? 1 : 0) +
dfs(node.left, sum - node.val) +
dfs(node.right, sum - node.val);
}
}

### Approach #2 Prefix Sum + backtrace

class Solution {
int count = 0;
int k;
HashMap<Integer, Integer> h = new HashMap();
public int pathSum(TreeNode root, int sum) {
k = sum;
preorder(root, 0);
return count;
}
public void preorder(TreeNode node, int currSum) {
if (node == null)
return;
// current prefix sum
currSum += node.val;
// here is the sum we're looking for
if (currSum == k)
count++;
// number of times the curr_sum − k has occured already,
// determines the number of times a path with sum k
// has occured upto the current node
count += h.getOrDefault(currSum - k, 0);
// add the current sum into hashmap
// to use it during the child nodes processing
h.put(currSum, h.getOrDefault(currSum, 0) + 1);
// process left subtree
preorder(node.left, currSum);
// process right subtree
preorder(node.right, currSum);
// remove the current sum from the hashmap
// in order not to use it during
// the parallel subtree processing
h.put(currSum, h.get(currSum) - 1);
}
}