### 437. Path Sum III

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``````

### 题解

1. `当前节点作为根节点`时相加为 sum 的路径之和。
2. `当前节点左子树作为根节点`时相加为 sum 的路径之和。
3. `当前节点右子树作为根节点`时相加为 sum 的路径之和。

``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} sum
* @return {number}
*/
var pathSum = function(root, sum) {
if (!root) return 0
// total means result value
// initialSum means init sum value
let result = { total: 0, initialSum: sum }
analyzeSum(root, result, sum)
return result.total
};

var analyzeSum = function(node, result, sum) {
if (!node) return
const extraSum = sum - node.val
if (extraSum === 0) {
result.total = result.total + 1
}

analyzeSum(node.left, result, extraSum)
analyzeSum(node.right, result, extraSum)
// handle it as root node.
analyzeSum(node.left, result, result.initialSum)
analyzeSum(node.right, result, result.initialSum)
}``````

submit 后, 卡在了如下测试用例中:

``````输入:
[1,null,2,null,3] // 该测试用例有点怪, 应该为 [1, null, 2, null, null, null, 3]
3

1
/   \
null    2
/   \
null    3

3

2``````

``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} sum
* @return {number}
*/
// recursive root node
var pathSum = function(root, sum) {
if (!root) return 0
const curTotal = analyzeSum(root, sum)
const leftChildTotal = pathSum(root.left, sum)
const rightChildTotal = pathSum(root.right, sum)
return curTotal + leftChildTotal + rightChildTotal
};

// recursive anlyze total value
var analyzeSum = function(node, expectSum) {
if (!node) return 0
const extraSum = expectSum - node.val
if (extraSum === 0) {
return 1 + analyzeSum(node.left, extraSum) + analyzeSum(node.right, extraSum)
} else {
return analyzeSum(node.left, extraSum) + analyzeSum(node.right, extraSum)
}
}``````