437. Path Sum III

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于 2019-03-16 12:04:01 发布

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本文探讨了在给定的二叉树中寻找路径数量，这些路径上的节点值之和等于特定目标值的问题。采用递归深度优先搜索(DFS)算法解决这一挑战，通过遍历树的每个节点并计算从该节点开始的路径是否达到指定和，从而找到所有可能的路径。

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题目描述

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.
在这里插入图片描述

方法思路

recursive DFS.

class Solution {
    //Runtime: 18 ms, faster than 33.29%
    //Memory Usage: 40.2 MB, less than 68.75%
    public int pathSum(TreeNode root, int sum) {
        if(root == null) return 0;
        int left = pathSum(root.left, sum);
        int right = pathSum(root.right, sum);
        return pathSumHelp(root, sum) + left + right;        
    }
    public int pathSumHelp(TreeNode root, int sum){
        if(root == null) return 0;
        int left = pathSumHelp(root.left, sum - root.val);
        int right = pathSumHelp(root.right, sum - root.val);
        return (root.val == sum ? 1 : 0) + left + right;
    }
}