LeetCode 437. Path Sum III

原题链接在这里：https://leetcode.com/problems/path-sum-iii/

题目：

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

题解：

采用DFS, DFS state needs current node and current sum.

DFS returns count of paths from current node.

PathSum returns dfs from current node + PathSum left child with sum, + PathSum right child with sum.  

Time Complexity: O(n^2). Space: O(logn), stack space.

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public 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);
    }
    
    private int dfs(TreeNode root, int sum){
        int res = 0;
        if(root == null){
            return res;
        }
        
        sum -= root.val;
        if(sum == 0){
            res++;
        }
        res += dfs(root.left, sum) + dfs(root.right, sum);
        return res;
    }
}

用HashMap来maintain prefix sum. Key 是prefix sum, value 是加到prefix sum的method count.

当cur - sum的值正好出现在prefix中时说明 cur-prefix正好是sum, 从prefix到cur的这一段加起来正好是sum.

Note: put (0L, 1) in the map at the beginning. Remember the backtracking, revert the map. The reason why map key is Long is because of integer overflow. e.g. Integer.MAX_VALUE + 1 as curSum.

Time Complexity: O(n). Space: O(n), hm.size().

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int pathSum(TreeNode root, int targetSum) {
        Map<Long, Integer> hm = new HashMap<>();
        hm.put(0L, 1); //设置prefix sum为0的default value是1.
        return dfs(root, targetSum, 0L, hm);
    }

    private int dfs(TreeNode root, int targetSum, long curSum, Map<Long, Integer> hm){
        if(root == null){
            return 0;
        }

        curSum += root.val;
        int res = hm.getOrDefault(curSum - targetSum, 0);
        hm.put(curSum, hm.getOrDefault(curSum, 0) + 1);
        res += dfs(root.left, targetSum, curSum, hm) + dfs(root.right, targetSum, curSum, hm);
        hm.put(curSum, hm.get(curSum) - 1); //remove count by 1, backtracking要算其他branch时先回复原值
        return res;
    }
}

类似Path Sum II, Path Sum IV.