[Leetcode] Path Sum

Path Sum 题解

题目来源：https://leetcode.com/problems/path-sum/description/

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Description

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Example

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Solution

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (!root)
            return false;
        if (root -> val == sum &&
                root -> left == NULL && root -> right == NULL)
            return true;
        return hasPathSum(root -> left, sum - (root -> val)) ||
                hasPathSum(root -> right, sum - (root -> val));
    }
};

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解题描述

这道题题意是对给出的数字sum，检索二叉树上根到叶子节点的路径，求路径上节点数据之和能否等于sum。上面给出的是递归实现DFS的解法，下面给出的是BFS的解法：

class Solution {
private:
    struct Task {
        TreeNode *node;
        int sum;
        Task(TreeNode* n, int s) : node(n), sum(s) {}
    };
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (!root)
            return false;
        queue<Task> q;
        q.push(Task(root, sum));
        while (!q.empty()) {
            Task task = q.front();
            q.pop();
            TreeNode *node = task.node;
            if (node -> left == NULL && node -> right == NULL &&
                    node -> val == task.sum)
                return true;
            if (node -> left) {
                q.push(Task(node -> left, task.sum - (node -> val)));
            }
            if (node -> right) {
                q.push(Task(node -> right, task.sum - (node -> val)));
            }
        }
        return false;
    }
};