• binary-tree-inorder-traversal


    /**
    *
    * @author gentleKay
    * Given a binary tree, return the inorder traversal of its nodes' values.
    * For example:
    * Given binary tree{1,#,2,3},
    1

    2
    /
    3
    * return[1,3,2].
    * Note: Recursive solution is trivial, could you do it iteratively?
    * confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
    * 给定二叉树,返回其节点值的中序遍历。
    * 例如:
    * 给定二叉树1,,2,3,
    * 返回[1,3,2]。
    * 注意:递归解决方案很简单,可以迭代吗?
    */

    推荐一个博客(关于递归和非递归二叉树的遍历)

    https://blog.csdn.net/wang454592297/article/details/79472938

    方法一:(非递归进行中序遍历)

    import java.util.ArrayList;
    import java.util.Stack;
    
    /**
     * 
     * @author gentleKay
     * Given a binary tree, return the inorder traversal of its nodes' values.
     * For example:
     * Given binary tree{1,#,2,3},
       1
        
         2
        /
       3
     * return[1,3,2].
     * Note: Recursive solution is trivial, could you do it iteratively?
     * confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
     * 给定二叉树,返回其节点值的中序遍历。
     * 例如:
     * 给定二叉树1,,2,3,
     * 返回[1,3,2]。
     * 注意:递归解决方案很简单,可以迭代吗?
     */
    
    public class Main21 {
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    //		TreeNode root = null;
    		TreeNode root = new TreeNode(4);
    		root.left = new TreeNode(2);
    		root.left.left = new TreeNode(1);
    		root.left.right  = new TreeNode(3);
    		
    		root.right = new TreeNode(6);
    		root.right.left = new TreeNode(5);
    		root.right.right = new TreeNode(7);
    		root.right.right.right = new TreeNode(8);
    		System.out.println(Main21.inorderTraversal(root));
    	}
    	
    	public static class TreeNode {
    		int val;
    		TreeNode left;
    		TreeNode right;
    		TreeNode(int x) { val = x; }
    	}
    	
    	public static ArrayList<Integer> inorderTraversal(TreeNode root) {
    		Stack<TreeNode> stack = new Stack<>();
    		ArrayList<Integer> array = new ArrayList<>();
    		while (root != null || !stack.isEmpty()) {
    			while (root != null) {
    				stack.push(root);
    				root = root.left;
    			}
    			if (!stack.isEmpty()) {
    				root = stack.pop();
    				array.add(root.val);
    				root = root.right;
    			}
    		}
    		return array;
    	}
    }
    

    方法二:(递归进行中序遍历)

    import java.util.ArrayList;
    import java.util.Stack;
    
    /**
     * 
     * @author gentleKay
     * Given a binary tree, return the inorder traversal of its nodes' values.
     * For example:
     * Given binary tree{1,#,2,3},
       1
        
         2
        /
       3
     * return[1,3,2].
     * Note: Recursive solution is trivial, could you do it iteratively?
     * confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
     * 给定二叉树,返回其节点值的中序遍历。
     * 例如:
     * 给定二叉树1,,2,3,
     * 返回[1,3,2]。
     * 注意:递归解决方案很简单,可以迭代吗?
     */
    
    public class Main21 {
    
    	public static void main(String[] args) {
    		// TODO Auto-generated method stub
    //		TreeNode root = null;
    		TreeNode root = new TreeNode(4);
    		root.left = new TreeNode(2);
    		root.left.left = new TreeNode(1);
    		root.left.right  = new TreeNode(3);
    		
    		root.right = new TreeNode(6);
    		root.right.left = new TreeNode(5);
    		root.right.right = new TreeNode(7);
    		root.right.right.right = new TreeNode(8);
    		System.out.println(Main21.inorderTraversal(root));
    	}
    	
    	public static class TreeNode {
    		int val;
    		TreeNode left;
    		TreeNode right;
    		TreeNode(int x) { val = x; }
    	}
    	
    	static ArrayList<Integer> array = new ArrayList<>();
    	public static ArrayList<Integer> inorderTraversal(TreeNode root) {
            if (root == null) {
            	return array;
            }
    		ergodic(root);
            return array;
        }
    	
    	public static void ergodic(TreeNode root) {
    		if (root.left != null) {
    			ergodic(root.left);
    		}
    		array.add(root.val);
    		if (root.right != null) {
    			ergodic(root.right);
    		}
    	}
    }
    

      

  • 相关阅读:
    nginx负载均衡
    mysqld: Out of memory Centos 创建swap分区解决
    redis 基本命令
    查看日志常用命令
    StringIO和BytesIO
    paramiko初识
    微信小程序-drf登录认证组件
    微信小程序之模块化--module.exports
    celery 定时任务报错一
    微信小程序跨页面传值
  • 原文地址:https://www.cnblogs.com/strive-19970713/p/11271596.html
Copyright © 2020-2023  润新知