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二叉树前中后序遍历

中序遍历链接:
https://leetcode-cn.com/problems/binary-tree-inorder-traversal/

后序遍历链接:
https://leetcode-cn.com/problems/binary-tree-postorder-traversal/

题目描述

题目分析

二叉树前中后序遍历主要提供了两种方法:

  • 迭代
  • 递归

前序遍历代码实现

public class PreOrder {


    public static void main(String[] args) {
        PreOrder preOrder = new PreOrder();
//        TreeNode root = new TreeNode(1, null, new TreeNode(2, new TreeNode(3), null));
        TreeNode root = null;
        preOrder.preorderTraversal(root);
        preOrder.preorder2(root);
    }


    /**
     * 迭代算法
     *
     * @param root
     * @return
     */
    public List<Integer> preorder2(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();

        if(root == null){
            return list;
        }

        LinkedList<TreeNode> stack = new LinkedList();
        stack.add(root);

        while (!stack.isEmpty()) {
            TreeNode treeNode = stack.getLast();
            list.add(treeNode.val);
            stack.removeLast();
            if (treeNode.right != null) {
                stack.add(treeNode.right);
            }
            if (treeNode.left != null) {
                stack.add(treeNode.left);
            }

        }

        System.out.println(list);
        return list;
    }


    /**
     * 递归
     *
     * @param root
     * @return
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();

        order(root, list);
        System.out.println(list);
        return list;
    }

    private void order(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        list.add(root.val);
        if (root.left != null) {
            order(root.left, list);
        }
        if (root.right != null) {
            order(root.right, list);
        }
    }


}

中序遍历代码实现

public class MiddleOrder {


    public static void main(String[] args) {
        MiddleOrder preOrder = new MiddleOrder();
        TreeNode root = new TreeNode(1, null, new TreeNode(2, new TreeNode(3), null));
//        TreeNode root = null;
        preOrder.preorderTraversal(root);
        preOrder.preorder2(root);
    }


    /**
     * 迭代算法
     *
     * @param root
     * @return
     */
    public List<Integer> preorder2(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();

        if (root == null) {
            return list;
        }
        LinkedList<TreeNode> stack = new LinkedList();

        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.addLast(root);
                root = root.left;
            }
            root = stack.removeLast();
            list.add(root.val);
            root = root.right;
        }

        System.out.println(list);
        return list;
    }


    /**
     * 递归
     *
     * @param root
     * @return
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();

        order(root, list);
        System.out.println(list);
        return list;
    }

    private void order(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        if (root.left != null) {
            order(root.left, list);
        }
        list.add(root.val);
        if (root.right != null) {
            order(root.right, list);
        }
    }


}

后序遍历代码实现

public class AfterOrder {


    public static void main(String[] args) {
        AfterOrder preOrder = new AfterOrder();
        TreeNode root = new TreeNode(1, null, new TreeNode(2, new TreeNode(3), null));
//        TreeNode root = null;
        preOrder.preorderTraversal(root);
        preOrder.preorder2(root);
    }


    /**
     * 迭代算法
     *
     * @param root
     * @return
     */
    public List<Integer> preorder2(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        if (root == null) {
            return list;
        }
        LinkedList<TreeNode> stack = new LinkedList();
        TreeNode prev = null;
        while (root != null || !stack.isEmpty()) {
            while (root != null) {
                stack.addLast(root);
                root = root.left;
            }
            root = stack.removeLast();
            if(root.right == null || root.right == prev){
                // 根节点
                list.add(root.val);
                prev = root;
                root = null;
            } else {
                stack.addLast(root);
                root = root.right;
            }
        }

        System.out.println(list);
        return list;
    }


    /**
     * 递归
     *
     * @param root
     * @return
     */
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();

        order(root, list);
        System.out.println(list);
        return list;
    }

    private void order(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        if (root.left != null) {
            order(root.left, list);
        }
        if (root.right != null) {
            order(root.right, list);
        }
        list.add(root.val);
    }


}

二叉树前中后序遍历,两种方法的复杂度

  • 时间复杂度:O(n)
  • 空间复杂度:O(n)

好了,今天就到这里,感谢各位看官到这里,不如点个关注吧!

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