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  • 路径总和

    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.

    Note: A leaf is a node with no children.

    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.



    题意:

    给定一个二叉树和一个目标和,判断该树中是否存在根节点到叶子节点的路径,使得路径上所有节点值相加等于目标和。

    思路:

    递归地调用判断函数,根据左右子树的具体情况进行判断分类。

    class Solution {
    public:
        bool hasPathSum(TreeNode* root, int sum) {   
            if (root == nullptr) return false;    
            return Traverse(root, 0, sum);
        }
        
        bool Traverse(TreeNode* root, int num, int target)
        {
            if (root == nullptr) return num == target;
            num += root->val;
            if (root->left != nullptr && root->right == nullptr)
            {
            return Traverse(root->left, num, target);
            }
            else if (root->right != nullptr && root->left == nullptr)
            {
            return Traverse(root->right, num, target);
            }
            else
            {
                return Traverse(root->left, num, target) ||
                Traverse(root->right, num, target);
            }
        }
    
    };
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  • 原文地址:https://www.cnblogs.com/Bipolard/p/9997081.html
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