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