Path Sum III (M)
題目
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:
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
題意
在給定的樹中找到一條從上到下的路徑，使其和為給定值，路徑的起點和終點可以是任意結點，
思路
雙重遞回，第一重遞回確定根結點，限制需要查找的子樹；第二重遞回在該子樹中找到從子樹根結點出發和為sum的路徑的個數，
代碼實作
Java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int pathSum(TreeNode root, int sum) {
if (root == null) {
return 0;
}
return findPath(root, 0, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
}
private int findPath(TreeNode root, int curSum, int sum) {
if (root == null) {
return 0;
}
curSum += root.val;
return (curSum == sum ? 1 : 0) + findPath(root.left, curSum, sum) + findPath(root.right, curSum, sum);
}
}
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