给定一个二叉树,它的每个结点都存放着一个整数值。
找出路径和等于给定数值的路径总数。
路径不需要从根节点开始,也不需要在叶子节点结束,但是路径方向必须是向下的(只能从父节点到子节点)。
二叉树不超过1000个节点,且节点数值范围是 [-1000000,1000000] 的整数。
示例:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8
10
/ \
5 -3
/ \ \
3 2 11
/ \ \
3 -2 1
返回 3。和等于 8 的路径有:
1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/path-sum-iii
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解法一:双递归
/**
* 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;
return pathSum(root.left,sum) + pathSum(root.right,sum) + count(root, sum, 0);
}
public int count(TreeNode root, int sum, int n) {
if(root == null) return 0;
n += root.val;
int m = 0;
if(sum == n) m++;
return count(root.left, sum, n) + count(root.right, sum, n) + m;
}
}
解法二:路径保存
/**
* 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;
int[] array = new int[1000];
return count(root, sum, array, 0);
}
public int count(TreeNode root, int sum, int[] array, int end) {
if(root == null) return 0;
int tmp = root.val;
int n = (tmp == sum)? 1 : 0;
for(int i = end-1; i >= 0; i--){
tmp += array[i];
if(sum == tmp) n++;
}
array[end] = root.val;
return count(root.left, sum, array, end+1) + count(root.right, sum, array, end+1) + n;
}
}