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437. Path Sum III - Easy

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

 

注意一下這裡的路徑不是從root到leaf,中間有部分的路徑加和為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; } }