437. Path Sum III-优快云博客

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/54944071

本文介绍了一种算法问题——给定一个二叉树及目标和，找出所有路径上节点值相加等于该目标和的路径数。路径不必从根节点开始或在叶子节点结束，但必须从父节点到子节点向下延伸。文中提供了两种解决方案，一种是递归方法，另一种使用了哈希映射来优化性能。

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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   1

Return 3. The paths that sum to 8 are:

1.  5 -> 3
2.  5 -> 2 -> 1
3. -3 -> 11

这里分别记录递归解法和hashmap解法：

递归解法，思路清晰，遍历各个子树，查找各个子路径，寻找sum=target的子路径，时间复杂度O(n^2)，代码如下：

/**
 * 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 findSub(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }
    public int findSub(TreeNode root, int sum) {
        int res = 0;
        if (root == null) {
            return 0;
        }
        if (sum == root.val) {
            res ++;
        }
        res += findSub(root.left, sum - root.val);
        res += findSub(root.right, sum - root.val);
        return res;
    }
}

Hashmap解法，时间复杂度O(n)，代码如下：

public int pathSum(TreeNode root, int sum) {
        HashMap<Integer, Integer> preSum = new HashMap();
        preSum.put(0,1);
        return helper(root, 0, sum, preSum);
    }
    
    public int helper(TreeNode root, int currSum, int target, HashMap<Integer, Integer> preSum) {
        if (root == null) {
            return 0;
        }
        
        currSum += root.val;
        int res = preSum.getOrDefault(currSum - target, 0);
        preSum.put(currSum, preSum.getOrDefault(currSum, 0) + 1);
        
        res += helper(root.left, currSum, target, preSum) + helper(root.right, currSum, target, preSum);
        preSum.put(currSum, preSum.get(currSum) - 1);
        return res;
    }