NowCoderTOP23-27 Binary tree traversal - continuous update ing

TOP23. 二叉树的前序遍历

public class Solution {
    
    /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * @param root TreeNode类 * @return int整型一维数组 */
    
    public void preorder(List<Integer> list, TreeNode root) {
    
        // 空节点就返回
        if(root == null) return;
        // 前序遍历 ——> 根 左 右
        list.add(root.val);
        preorder(list, root.left);
        preorder(list, root.right);
    } 
    
    public int[] preorderTraversal (TreeNode root) {
    
        // The final result is an array,so先定义一个数组
        List<Integer> list = new ArrayList();
        preorder(list, root);
        // 定义一个数组长度为list大小
        int[] res = new int[list.size()];
        for(int i = 0; i < list.size(); i++) {
    
            res[i] = list.get(i);
        }
        return res;
    }
}

TOP24. 二叉树的中序遍历

public class Solution {
    
    /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * @param root TreeNode类 * @return int整型一维数组 */
    public void inorder(List<Integer> list, TreeNode root) {
    
        // 先判断树是否为空
        if(root == null) return;
        // 中序遍历 ——> 左 根 右
        inorder(list, root.left);
        list.add(root.val);
        inorder(list, root.right);
    }
    
    public int[] inorderTraversal (TreeNode root) {
    
        List<Integer> list = new ArrayList();
        inorder(list, root);
        // Define an array to store the results of in-order traversal
        int[] res = new int[list.size()];
        for(int i = 0; i < list.size(); i++) {
    
            res[i] = list.get(i);
        }
        return res;
    }
}

TOP25. 二叉树的后序遍历

public class Solution {
    
    /** * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 * @param root TreeNode类 * @return int整型一维数组 */
    public void postorder(List<Integer> list, TreeNode root) {
    
        if(root == null) return;
        postorder(list, root.left);
        postorder(list, root.right);
        list.add(root.val);
    }
    
    public int[] postorderTraversal (TreeNode root) {
    
        List<Integer> list = new ArrayList();
        postorder(list, root);
        int[] res = new int[list.size()];
        for(int i = 0; i < list.size(); i++) {
    
            res[i] = list.get(i);
        }
        return res;
    }
}

TOP26. 求二叉树的层序遍历

public class Solution {
    
    /** * 使用队列实现层序遍历 * @param root TreeNode类 * @return int整型ArrayList<ArrayList<>> */
    public ArrayList<ArrayList<Integer>> levelOrder (TreeNode root) {
    
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>> ();
        if(root == null) return res;
        // 队列存储,进行层次遍历
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        // 根节点先入队
        queue.offer(root);
        // 当队列不为空时
        while(!queue.isEmpty()){
    
            // 记录二叉树的某一行
            ArrayList<Integer> row = new ArrayList<Integer>();
            // The size of the queue is the number of elements in this layer
            int n = queue.size();
            // Start traversing all elements of this layer
            for(int i = 0; i < n; i++) {
    
                TreeNode cur = queue.poll();
                row.add(cur.val);
                //若是左右孩子存在,则存入左右孩子作为下一个层次
                if(cur.left != null) queue.add(cur.left);
                if(cur.right != null) queue.add(cur.right);
            }
            // Bring all nodes of a layer into the total result set
            res.add(row);
        }
        return res;
    }
}

TOP27. 按之字形顺序打印二叉树

public class Solution {
    
    public ArrayList<ArrayList<Integer> > Print(TreeNode pRoot) {
    
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>> ();
        if(pRoot == null) return res;
        // 队列存储,进行层次遍历
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        // 根节点先入队
        queue.offer(pRoot);
        // 标记位,行数为偶数,逆置
        Boolean flag = true;
        // 当队列不为空时
        while(!queue.isEmpty()){
    
            flag = !flag;
            // 记录二叉树的某一行
            ArrayList<Integer> row = new ArrayList<Integer>();
            // The size of the queue is the number of elements in this layer
            int n = queue.size();
            // Start traversing all elements of this layer
            for(int i = 0; i < n; i++) {
    
                TreeNode cur = queue.poll();
                row.add(cur.val);
                //若是左右孩子存在,则存入左右孩子作为下一个层次
                if(cur.left != null) queue.add(cur.left);
                if(cur.right != null) queue.add(cur.right);
            }
            if (flag)
            {
    
                Collections.reverse(row);
            }
            // Bring all nodes of a layer into the total result set
            res.add(row);
        }
        return res;
    }
}