Problem 437: Path Sum III
思路
这道题相当于是两部分的递归,第一部分递归这个树,对于每个 root,设计到了第二部分的递归 -- 求 count 数
这两部分的递归一定要先想清楚,否则主函数的 return 很容易写错
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int pathSum(TreeNode root, int sum) {
if (root == null) return 0;
return countPath(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
}
private int countPath(TreeNode root, int sum) {
if (root == null) return 0;
int count = 0;
if (root.val == sum) count++;
count += countPath(root.left, sum - root.val);
count += countPath(root.right, sum - root.val);
return count;
}
}
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