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

这道题是要找到满足sum == target的路径个数，路径不一定是root to leaf, 但是必须是从上到下的。注意看一开始的错误解法。

submit 1 : int基本类型参数传入的是值，一旦传入是不会像传入引用类型的地址值一样， 在函数运行过程中地址所指向的值会改变的。所以你的res = 0传到函数里，res的值永远不会改变，一直是0.

image.png

submit 2 : 边界条件错误， for循环执行的条件是边界条件为true,跟while一样。所以这里你表达的意思应该是i >= 0,而不是i== 0, 后者作为边界条件时for循环根本不会执行。

53D7D469-5DB1-46DA-9151-9D35E1BC5AF2.png

/**
 * 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) {
        List path = new ArrayList<>();
        return helper(root, path, sum); 
    }
    
    private int helper(TreeNode root, List path, int sum){
        int res = 0;
        if (root == null){
            return res;
        }
        path.add(root.val);
        int pathSum = 0;
        for (int i = path.size() - 1; i >= 0; i--){
            pathSum += path.get(i);
            if (pathSum == sum){
                res++;
            }
        }
        res += helper(root.left, path, sum);
        res += helper(root.right, path, sum);
        path.remove(path.size() - 1);
        return res;
    }
}