[LeetCode] 437. Path Sum III

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

路径和III。题意和前两个版本类似，但是这个题找的路径和，起点可以不是根节点，问满足路径和为sum的路径有多少。

这个题出的非常好。思路是前缀和。利用前缀和的思路，递归统计你遍历到的子树上的前缀和presum是多少，并将这个前缀和存入hashmap。其他的部分就是很常规的前序遍历。

时间O(n)

空间O(n)

Java实现

/**
 * 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 sum) {
        HashMap<Integer, Integer> map = new HashMap();
        map.put(0, 1);
        helper(root, 0, sum, map);
        return count;
    }
    
    int count = 0;
    public void helper(TreeNode root, int curSum, int target, HashMap<Integer, Integer> map) {
        // corner case
        if (root == null) {
            return;
        }
        
        // normal case
        curSum += root.val;
        if (map.containsKey(curSum - target)) {
            count += map.get(curSum - target);
        }
        if (!map.containsKey(curSum)) {
            map.put(curSum, 1);
        } else {
            map.put(curSum, map.get(curSum) + 1);
        }
        helper(root.left, curSum, target, map);
        helper(root.right, curSum, target, map);
        map.put(curSum, map.get(curSum) - 1);
    }
}

前缀和prefix sum题目总结

LeetCode 题目总结