栈和队列

20. 有效的括号

class Solution {
    public boolean isValid(String s) {
        Map<Character, Character> map = new HashMap<>();
        map.put('(', ')');
        map.put('{', '}');
        map.put('[', ']');
        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
            Character ch = s.charAt(i);
            if (ch == '(' || ch == '{' || ch == '[') {
                stack.push(ch);
            } else if (stack.isEmpty() || ch != map.get(stack.pop())) {
                return false;
            }
        }
        return stack.isEmpty();
    }
}

https://leetcode.cn/problems/valid-parentheses

144. 二叉树的前序遍历

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        // 递归方式
        if (root == null) {
            return Collections.emptyList();
        }
        List<Integer> list = new ArrayList<>();
        list.add(root.val);
        list.addAll(preorderTraversal(root.left));
        list.addAll(preorderTraversal(root.right));
        return list;
    }
}

class Solution {
    private List<Integer> list = new ArrayList<>();
    public List<Integer> preorderTraversal(TreeNode root) {
        // 非递归方式
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while (node != null || !stack.empty()) {
            if (node != null) {
                stack.push(node);
                list.add(node.val);
                node = node.left;
            } else {
                node = stack.pop().right;
            }
        }
        return list;
    }
}

https://leetcode.cn/problems/binary-tree-preorder-traversal

94. 二叉树的中序遍历

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        // 递归方式
        if (root == null) {
            return Collections.emptyList();
        }
        List<Integer> list = new ArrayList<>();
        list.addAll(inorderTraversal(root.left));
        list.add(root.val);
        list.addAll(inorderTraversal(root.right));
        return list;
    }
}

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        // 非递归方式
        List<Integer> result = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while (node != null || !stack.empty()) {
            if (node != null) {
                stack.push(node);
                node = node.left;
            } else {
                TreeNode top = stack.pop();
                result.add(top.val);
                node = top.right;
            }
        }
        return result;
    }
}

https://leetcode.cn/problems/binary-tree-inorder-traversal

145. 二叉树的后序遍历

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        // 递归方式
        if (root == null) {
            return Collections.emptyList();
        }
        List<Integer> list = new ArrayList<>();
        list.addAll(postorderTraversal(root.left));
        list.addAll(postorderTraversal(root.right));
        list.add(root.val);
        return list;
    }
}

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        // 非递归方式
        List<Integer> result = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        while (node != null || !stack.empty()) {
            if (node != null) {
                stack.push(node);
                result.add(node.val);
                node = node.right;
            } else {
                node = stack.pop().left;
            }
        }
        Collections.reverse(result);
        return result;
    }
}

https://leetcode.cn/problems/binary-tree-postorder-traversal

102. 二叉树的层序遍历

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        if (root == null) {
            return Collections.emptyList();
        }
        List<List<Integer>> result = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                list.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            result.add(list);
        }
        return result;
    }
}

https://leetcode.cn/problems/binary-tree-level-order-traversal

347. 前K个高频元素

class Solution {
    public int[] topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int num : nums) {
            freq.compute(num, (kk, v) -> v == null ? 1 : v + 1);
        }

        PriorityQueue<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>(
            (e1, e2) -> {
                return e1.getValue() - e2.getValue();
            }
        );
        for (Map.Entry<Integer, Integer> entry : freq.entrySet()) {
            if (queue.size() == k) {
                if (queue.peek().getValue() < entry.getValue()) {
                    queue.poll();
                    queue.offer(entry);
                }
            } else {
                queue.offer(entry);
            }
            
        }

        int[] result = new int[k];
        int i = 0;
        while(!queue.isEmpty()) {
            result[i++] = queue.poll().getKey();
        }
        return result;
    }
}

https://leetcode.cn/problems/top-k-frequent-elements