437. Path Sum III(统计路径和等于一个数的路径数量)
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
路径不一定以 root 开头,也不一定以 leaf 结尾,但是必须连续。
方法一:递归
时间复杂度:o(n) 空间复杂度:o(1)
/** * 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 ret=0;
//累加,每次从根节点开始迭代,找到了就返回1,找到叶子结点还找不到,就返回0。下次从根节点的叶子结点找,依次递归遍历所有可能的组合。 ret=pathSumStartwithRoot(root,sum)+pathSum(root.left,sum)+pathSum(root.right,sum); return ret; } private int pathSumStartwithRoot(TreeNode root, int sum){ //这个地方思路和112题类似 if(root==null) return 0; int ret=0; if(root.val==sum) ret++; ret+=pathSumStartwithRoot(root.left,sum-root.val)+pathSumStartwithRoot(root.right,sum-root.val); return ret; } }
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