本代码提交于 2021-2-10 15:02 (本项目第二次提交)

Signed-off-by: icbug-docs <azmicbug@outlook.com>

Author: icbug

111.二叉树的最小深度.cpp

@@ -0,0 +1,22 @@
+class Solution {
+public:
+    int minDepth(TreeNode *root) {
+        if (root == nullptr) {
+            return 0;
+        }
+
+        if (root->left == nullptr && root->right == nullptr) {
+            return 1;
+        }
+
+        int min_depth = INT_MAX;
+        if (root->left != nullptr) {
+            min_depth = min(minDepth(root->left), min_depth);
+        }
+        if (root->right != nullptr) {
+            min_depth = min(minDepth(root->right), min_depth);
+        }
+
+        return min_depth + 1;
+    }
+};

112.路径总和.cpp

@@ -0,0 +1,31 @@
+class Solution {
+public:
+    bool hasPathSum(TreeNode *root, int sum) {
+        if (root == nullptr) {
+            return false;
+        }
+        queue<TreeNode *> que_node;
+        queue<int> que_val;
+        que_node.push(root);
+        que_val.push(root->val);
+        while (!que_node.empty()) {
+            TreeNode *now = que_node.front();
+            int temp = que_val.front();
+            que_node.pop();
+            que_val.pop();
+            if (now->left == nullptr && now->right == nullptr) {
+                if (temp == sum) return true;
+                continue;
+            }
+            if (now->left != nullptr) {
+                que_node.push(now->left);
+                que_val.push(now->left->val + temp);
+            }
+            if (now->right != nullptr) {
+                que_node.push(now->right);
+                que_val.push(now->right->val + temp);
+            }
+        }
+        return false;
+    }
+};

113. 路径总和 II.cpp

@@ -0,0 +1,24 @@
+class Solution {
+public:
+    vector<vector<int>> ret;
+    vector<int> path;
+
+    void dfs(TreeNode* root, int sum) {
+        if (root == nullptr) {
+            return;
+        }
+        path.emplace_back(root->val);
+        sum -= root->val;
+        if (root->left == nullptr && root->right == nullptr && sum == 0) {
+            ret.emplace_back(path);
+        }
+        dfs(root->left, sum);
+        dfs(root->right, sum);
+        path.pop_back();
+    }
+
+    vector<vector<int>> pathSum(TreeNode* root, int sum) {
+        dfs(root, sum);
+        return ret;
+    }
+};

114. 二叉树展开为链表.java

@@ -0,0 +1,20 @@
+class Solution {
+    public void flatten(TreeNode root) {
+        List<TreeNode> list = new ArrayList<TreeNode>();
+        preorderTraversal(root, list);
+        int size = list.size();
+        for (int i = 1; i < size; i++) {
+            TreeNode prev = list.get(i - 1), curr = list.get(i);
+            prev.left = null;
+            prev.right = curr;
+        }
+    }
+
+    public void preorderTraversal(TreeNode root, List<TreeNode> list) {
+        if (root != null) {
+            list.add(root);
+            preorderTraversal(root.left, list);
+            preorderTraversal(root.right, list);
+        }
+    }
+}

115. 不同的子序列.java

@@ -0,0 +1,13 @@
+class Solution {
+    public int numDistinct(String s, String t) {
+        int[][] dp = new int[t.length() + 1][s.length() + 1];
+        for (int j = 0; j < s.length() + 1; j++) dp[0][j] = 1;
+        for (int i = 1; i < t.length() + 1; i++) {
+            for (int j = 1; j < s.length() + 1; j++) {
+                if (t.charAt(i - 1) == s.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1];
+                else dp[i][j] = dp[i][j - 1];
+            }
+        }
+        return dp[t.length()][s.length()];
+    }
+}

116. 填充每个节点的下一个右侧节点指针.java

@@ -0,0 +1,41 @@
+class Solution {
+    public Node connect(Node root) {
+        if (root == null) {
+            return root;
+        }
+        
+        // 初始化队列同时将第一层节点加入队列中，即根节点
+        Queue<Node> queue = new LinkedList<Node>(); 
+        queue.add(root);
+        
+        // 外层的 while 循环迭代的是层数
+        while (!queue.isEmpty()) {
+            
+            // 记录当前队列大小
+            int size = queue.size();
+            
+            // 遍历这一层的所有节点
+            for (int i = 0; i < size; i++) {
+                
+                // 从队首取出元素
+                Node node = queue.poll();
+                
+                // 连接
+                if (i < size - 1) {
+                    node.next = queue.peek();
+                }
+                
+                // 拓展下一层节点
+                if (node.left != null) {
+                    queue.add(node.left);
+                }
+                if (node.right != null) {
+                    queue.add(node.right);
+                }
+            }
+        }
+        
+        // 返回根节点
+        return root;
+    }
+}

117. 填充每个节点的下一个右侧节点指针 II.java

@@ -0,0 +1,44 @@
+// package com.leetcode.explore.learnCard.dataStructureBinaryTree.conclusion4;
+
+/**
+ * 补充节点的右侧指针，不是完美二叉树
+ */
+public class Solution {
+    public Node connect(Node root) {
+        if (root == null || (root.right == null && root.left == null)) {
+            return root;
+        }
+        if (root.left != null && root.right != null) {
+            root.left.next = root.right;
+            root.right.next = getNextNoNullChild(root);
+        }
+        if (root.left == null) {
+            root.right.next = getNextNoNullChild(root);
+        }
+        if (root.right == null) {
+            root.left.next = getNextNoNullChild(root);
+        }
+
+        //这里要注意：先递归右子树，否则右子树根节点next关系没建立好，左子树到右子树子节点无法正确挂载
+        root.right = connect(root.right);
+        root.left = connect(root.left);
+
+        return root;
+    }
+
+    /**
+     * 一路向右找到有子节点的根节点
+     */
+    private static Node getNextNoNullChild(Node root) {
+        while (root.next != null) {
+            if (root.next.left != null) {
+                return root.next.left;
+            }
+            if (root.next.right != null) {
+                return root.next.right;
+            }
+            root = root.next;
+        }
+        return null;
+    }
+}

118.杨辉三角.cpp

@@ -0,0 +1,14 @@
+class Solution {
+public:
+    vector<vector<int>> generate(int numRows) {
+        vector<vector<int>> ret(numRows);
+        for (int i = 0; i < numRows; ++i) {
+            ret[i].resize(i + 1);
+            ret[i][0] = ret[i][i] = 1;
+            for (int j = 1; j < i; ++j) {
+                ret[i][j] = ret[i - 1][j] + ret[i - 1][j - 1];
+            }
+        }
+        return ret;
+    }
+};

119. 杨辉三角 II(进阶版).cpp

@@ -0,0 +1,32 @@
+class Solution1 {
+public:
+    vector<int> getRow(int rowIndex) {
+        vector<vector<int>> dp(rowIndex + 1, vector<int>(rowIndex + 1, 0));
+        for (int i = 0; i <= rowIndex; i++) {
+            dp[i][0] = 1;
+            dp[i][i] = 1;
+        }
+
+        for (int i = 1; i <= rowIndex; i++) {
+            for (int j = 1; j < i; j++) {
+                dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
+            }
+        }
+
+        return dp[rowIndex];
+    }
+};
+
+class Solution {
+public:
+    vector<int> getRow(int rowIndex) {
+        vector<int> dp(rowIndex + 1, 1);
+        for (int i = 1; i <= rowIndex; i++) {
+            for (int j = i - 1; j >= 1; j--) {
+                dp[j] = dp[j - 1] + dp[j];
+            }
+        }
+
+        return dp;
+    }
+};

120.三角形最最短路径和.cpp

@@ -0,0 +1,16 @@
+class Solution {
+public:
+    int minimumTotal(vector<vector<int>>& triangle) {
+        int n = triangle.size();
+        vector<vector<int>> f(n, vector<int>(n));
+        f[0][0] = triangle[0][0];
+        for (int i = 1; i < n; ++i) {
+            f[i][0] = f[i - 1][0] + triangle[i][0];
+            for (int j = 1; j < i; ++j) {
+                f[i][j] = min(f[i - 1][j - 1], f[i - 1][j]) + triangle[i][j];
+            }
+            f[i][i] = f[i - 1][i - 1] + triangle[i][i];
+        }
+        return *min_element(f[n - 1].begin(), f[n - 1].end());
+    }
+};