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题目: 解答: /* // Definition for a Node. class Node { public: int val; Node* left; Node* right; Node* next; Node() : val(0), left(NULL), right(NULL), next 阅读全文
posted @ 2020-05-02 13:11
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题目: 解法: 1 /* 2 // Definition for a Node. 3 class Node { 4 public: 5 int val; 6 Node* left; 7 Node* right; 8 Node* next; 9 10 Node() : val(0), left(NUL 阅读全文
posted @ 2020-05-02 13:07
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题目: 解答: 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode() : val( 阅读全文
posted @ 2020-05-02 13:04
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题目: 解法: 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : 阅读全文
posted @ 2020-05-02 13:01
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题目: 解答: 声明一个变量记录已经经过的节点的值之和,每经过一个节点就加上这个节点的值,在叶子节点判断变量值是否为目标值。 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * Tre 阅读全文
posted @ 2020-05-02 12:59
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题目: 解答: 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : 阅读全文
posted @ 2020-05-02 12:52
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题目: 解答: 平衡二叉树要求左子树和右子树的高度相差为1,且左右子树都是平衡二叉树,显然要计算二叉树的高度的函数。 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNod 阅读全文
posted @ 2020-05-02 12:46
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题目: 解答: 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : 阅读全文
posted @ 2020-05-02 12:40
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题目: 解答: 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : 阅读全文
posted @ 2020-05-02 12:35
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题目: 解答: 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : 阅读全文
posted @ 2020-05-02 12:32
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