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; } }

 

转载于:https://www.cnblogs.com/shaer/p/10570744.html

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