437. Path Sum III - Easy

发布时间：2020-12-14 05:17:53 所属栏目：大数据来源：网络整理

导读：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

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.

Example:

root = [10,5,-3,3,2,null,11,-2,1],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

注意一下这里的路径不是从root到leaf，中间有部分的路径加和为sum也可

time: O(n),space: O(height)

/**
 * 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;
        }
        return dfs(root,sum) + pathSum(root.left,sum) + pathSum(root.right,sum);
    }
    
    public int dfs(TreeNode root,int sum) {
        if(root == null) {
            return 0;
        }
        int cnt = 0;
        if(root.val == sum) {
            cnt++;
        }
        cnt += dfs(root.left,sum - root.val);
        cnt += dfs(root.right,sum - root.val);
        return cnt;
    }
}

（编辑：李大同）

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