leetcode:112 Path Sum－每日编程第二十七题-优快云博客

本文深入解析路径总和问题，通过实例演示如何利用递归算法解决二叉树中从根节点到叶节点的路径和等于给定数值的问题。包括问题背景、解题思路、代码实现及示例解析。

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Path Sum

Total Accepted: 81534 Total Submissions: 268857 Difficulty: Easy

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.

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.

思路：

注意val可能为正，可能为负。递归即可解决。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum,const int& expected){
        if(root==NULL){
            return false;
        }
        sum+=root->val;
        if(sum==expected){
            if(root->left==NULL&&root->right==NULL){
                return true;
            }
        }
        return hasPathSum(root->left,sum,expected)||hasPathSum(root->right,sum,expected);
    }

    bool hasPathSum(TreeNode* root, int sum) {
        if(root==NULL){
            return false;
        }
        return hasPathSum(root, 0, sum);
    }
};