LeetCode 437 Path Sum III (DFS)

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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   1Return 3. The paths that sum to 8 are:1.  5 -> 32.  5 -> 2 -> 13. -3 -> 11

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题目链接：https://leetcode.com/problems/path-sum-iii/

题目分析：因为只能向下，因此直接DFS即可，用个flag标记当前点是否可作为某次的起点

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {        void DFS(TreeNode root, boolean flag, int cur, int sum, int[] ans) {        cur += root.val;        if (cur == sum) {            ans[0] ++;        }        if (root.left != null) {            DFS(root.left, false, cur, sum, ans);            if (flag) {                DFS(root.left, true, 0, sum, ans);            }        }        if (root.right != null) {            DFS(root.right, false, cur, sum, ans);            if (flag) {                DFS(root.right, true, 0, sum, ans);            }        }    }        public int pathSum(TreeNode root, int sum) {        if (root == null) {            return 0;        }        int[] ans = new int[1];        DFS(root, true, 0, sum, ans);        return ans[0];    }}

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