/**
*
* @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); } } }