Problem: 437. 路径总和 III

题目描述

思路

1.定义int类型函数rootSum(root, targetSum)，用于求取每一个节点等于目标函数的路径数：

1.1.易知rootSum(root, targetSum)求出的数量等于rootSum(root.left, targetSum - value)求出的数量加上rootSum(root.right, targetSum - value)求出的数量,其中value是当前遍历到的节点的节点值
1.2.若当前的value值等于targetSum，则所求的路径总和加一；

2.在pathSum函数中实现对每一个节点调用rootSum函数得出最终的路劲总和数

复杂度

时间复杂度:

$O(n^2)$;其中$n$为二叉树节点的个数

空间复杂度:

$O(n)$

Code

/*** 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 {/*** Path Sum III** @param root      The root of binary tree* @param targetSum The target number* @return int*/public int pathSum(TreeNode root, long targetSum) {if (root == null) {return 0;}int res = rootSum(root, targetSum);res += pathSum(root.left, targetSum);res += pathSum(root.right, targetSum);return res;}/*** Find the sum of the target paths of each node** @param root      The root of binary tree* @param targetSum The target number* @return int*/public int rootSum(TreeNode root, long targetSum) {int res = 0;if (root == null) {return 0;}int value = root.val;if (value == targetSum) {res++;}res += rootSum(root.left, targetSum - value);res += rootSum(root.right, targetSum - value);return res;}
}