LeetCode 111. 二叉树的最小深度
分析
递归分析左右子树.
根据定义, 如果u是叶子结点, 那么深度为1.
如果不是叶子节点, 分成3种情况
1.左子树a, 右子树b 不空, 返回min(f(a), f(b)) + 1
2.a不空, b空, 返回f(a) + 1
3.a空, b不空 返回f(b) + 1
代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if (!root) return 0;
if (!root->left && !root->right) return 1; // 如果是叶子节点 返回1
if (root->left && root->right) return min(minDepth(root->left), minDepth(root->right)) + 1;
if (root->left) return minDepth(root->left) + 1;
return minDepth(root->right) + 1;
}
};
LeetCode 112. 路径总和
分析
自上而下递归, 然后判断叶子结点的sum是否==0
代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
if (!root) return false;
sum -= root->val;
if (!root->left && !root->right) return !sum;
return root->left && hasPathSum(root->left, sum) || root->right && hasPathSum(root->right, sum);
}
};
LeetCode 113. 路径总和 II
分析
新建一个dfs函数去递归, 不要在原函数中递归
还要新建全局变量res, path
代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> res;
vector<int> path;
vector<vector<int>> pathSum(TreeNode* root, int sum) {
if (root) dfs(root, sum);
return res;
}
void dfs(TreeNode* root, int sum){
path.push_back(root->val);
sum -= root->val;
if (!root->left && !root->right){
if (!sum) res.push_back(path);
}else{
if (root->left) dfs(root->left, sum);
if (root->right) dfs(root->right, sum);
}
path.pop_back();
}
};
LeetCode 114. 二叉树展开为链表
分析
代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void flatten(TreeNode* root) {
while (root){
auto p = root->left;
if (p){
while (p->right) p = p->right;
p->right = root->right;
root->right = root->left;
root->left = NULL;
}
root = root->right;
}
}
};
LeetCode 115. 不同的子序列
分析
闫氏dp分析法.
f[i][j] 表示s[1~i]所有前i个字符和t前j个字符匹配的子序列
注意 右边是无条件可以选择的情况, 所以f[i][j] = f[i - 1][j]
中间结果会爆int, 用longlong存储
代码
class Solution {
public:
int numDistinct(string s, string t) {
int n = s.size(), m = t.size();
s = ' ' + s, t = ' ' + t;
vector<vector<long long >> f(n + 1, vector<long long>(m + 1));
for (int i = 0; i < n; i ++ ) f[i][0] = 1; // 注意i从0开始
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= m; j ++ ){
f[i][j] = f[i - 1][j];
if (s[i] == t[j]) f[i][j] += f[i - 1][j - 1];
}
return f[n][m];
}
};