[LeetCode] Path Sum, Solution

最新推荐文章于 2020-07-07 13:07:47 发布

原创最新推荐文章于 2020-07-07 13:07:47 发布 · 240 阅读

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本文探讨了一道经典的二叉树遍历问题：给定一棵二叉树及一个目标和，判断是否存在从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定的目标和。通过递归方法解决此问题。

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.

» Solve this problem

[Thoughts]
二叉树遍历。遍历过程中累加节点值，当到达任意叶节点的时候，进行判断。

[Code]

1:       bool hasPathSum(TreeNode *root, int sum) {  
2:            return hasPathSum(root, 0, sum);      
3:       }  
4:       bool hasPathSum(TreeNode *root, int sum, int target) {  
5:            if(root == NULL) return false;  
6:            sum += root->val;  
7:            if(root->left == NULL && root->right == NULL) //leaf  
8:            {  
9:                 if(sum == target)  
10:                      return true;  
11:                 else  
12:                      return false;  
13:            }  
14:            return hasPathSum(root->left, sum, target)   
15:                   || hasPathSum(root->right, sum, target);      
16:       }