Leetcode: mimimum depth of tree, path sum, path sum II

思路:

简单搜索

总结:

dfs 框架

1. 需要打印路径. 在 dfs 函数中假如 vector 变量, 不用 & 修饰的话就不需要 undo

2. 不需要打印路径, 可设置全局变量 ans, 在 dfs 函数中对 ans 判定, 判定的位置尽可能的多

3. 对 tree 遍历, 有两种办法, 第一种是 if(root == NULL) 第二种是 if(root->left == NULL), 我一般用第二种, 效率比较高, 但是在第二种 return 1, 第一种 return 0

4. Leetcode 给出的测试用例经常会有空的输入, 要注意

5. path sum 中树节点的 val 可以是负数, 使得剪枝变得比较困难. 这个地方 wa 过

6. path sum II 题目没要求去重, 去重的话可能比较复杂, 我暂时没想到好办法

7. vector.clear() 可以彻底清空 vector, 不需要 for 循环

代码:

minimum depth of tree

 const int INF = 1E9;
class Solution {
public:int minDepth(TreeNode *root) {if(root == NULL)return 0;if(root->left == NULL && root->right == NULL) return 1;int left = INF, right = INF;if(root->left) {left = 1 + minDepth(root->left);         }if(root->right)right = 1 + minDepth(root->right);return min(left, right);}
};

path sum

class Solution {
public:int SUM;bool ans;void dfs(TreeNode *root, const int &curSum) {if(ans)return;if(curSum + root->val == SUM) {if(root->left == NULL && root->right == NULL) {ans = true;return;}}if(root->left != NULL && !ans) {dfs(root->left, curSum+root->val);}if(root->right != NULL && !ans) {dfs(root->right, curSum+root->val);}}bool hasPathSum(TreeNode *root, int sum) {SUM = sum;ans = false;if(root == NULL)return false;elsedfs(root, 0);return ans;}
};

path sum II

class Solution {
public:int SUM;vector<vector<int> > result;void dfs(TreeNode *root, const int &curSum, vector<int> party) {party.push_back(root->val);if(curSum + root->val == SUM) {if(root->left == NULL && root->right == NULL) {result.push_back(party);return;}}if(root->left != NULL ) {dfs(root->left, curSum+root->val, party);}if(root->right != NULL ) {dfs(root->right, curSum+root->val, party);}}vector<vector<int> > pathSum(TreeNode *root, int sum) {SUM = sum;result.clear();if(root != NULL) {vector<int> temp;dfs(root, 0, temp);}return result;}
};

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