437. Path Sum III - Easy

标签：range   ==   more   col   val   number   contain   return   efi   

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

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;
    }
}

437. Path Sum III - Easy

标签：range   ==   more   col   val   number   contain   return   efi   

原文地址：https://www.cnblogs.com/fatttcat/p/10198793.html