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LeetCode112. Path Sum(二叉树 / 搜索)

2019 年 03 月 15 日 • 阅读: 758 • LeetCode阅读设置

LeetCode112. Path Sum(二叉树 / 搜索)

  • Easy
  • Accepted:290,680
  • Submissions:782,733

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.


链接

https://leetcode.com/problems/path-sum

题意

给定一颗二叉树和一个数sum,判断这棵树是否存在一条从根结点到叶子结点的路径,该路径上所有结点的值加起来等于sum。

题解

对于当前结点root和sum值,我们可以判断到当前结点时是否已经符合条件,即判断结点值是否等于sum且该结点为叶子结点(即左右子树均为空),如果不符合条件,我们就接着递归判断该结点的左右子树,但是要注意此时的sum值需要减去当前结点值。

对每个结点访问一次,平均时间复杂度$O(n^2)$。

代码

  • Runtime: 16 ms, faster than 99.96% of C++ online submissions for Path Sum.
  • Memory Usage: 20 MB, less than 14.90% of C++ online submissions for Path Sum.
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (root == NULL)
            return false;
        if (root->val == sum && root->left == NULL && root->right == NULL) // 当前结点是否符合条件
            return true;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); // 递归判断左右子树
    }
};

The end.
2019年3月15日 星期五