Java写题解 | #实现二叉树先序,中序和后序遍历#
实现二叉树先序,中序和后序遍历
http://www.nowcoder.com/practice/a9fec6c46a684ad5a3abd4e365a9d362
二叉树的前序中序后序遍历 NC 45 LC 144 LC 94 LC 145
概述
前序遍历:根节点、左子树、右子树
中序遍历:左子树、根节点、右子树
后序遍历:左子树、右子树、根节点
主要有三种遍历方法:递归、迭代、Morris遍历
递归方法:
时间复杂度:
空间复杂度:平均情况为 最坏情况为 (即树呈链状)
迭代方法:
即显式地维护递归方法中的栈
时间复杂度、空间复杂度和递归方法一致。
Morris遍历:
时间复杂度:
空间复杂度:
(代码可参考leetcode题解,等掌握了再更新...)
1. 递归遍历
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * } */ public class Solution { /** * * @param root TreeNode类 the root of binary tree * @return int整型二维数组 */ List<Integer> preOrder; List<Integer> inOrder; List<Integer> postOrder; public int[][] threeOrders (TreeNode root) { // write code here preOrder = new ArrayList<>(); inOrder = new ArrayList<>(); postOrder = new ArrayList<>(); setPreOder(root); setInOrder(root); setPostOrder(root); int[][] ans = new int[3][preOrder.size()]; for (int i = 0; i < preOrder.size(); i++) { ans[0][i] = preOrder.get(i).intValue(); ans[1][i] = inOrder.get(i).intValue(); ans[2][i] = postOrder.get(i).intValue(); } return ans; } private void setPreOder(TreeNode root) { if (root != null) { preOrder.add(root.val); setPreOder(root.left); setPreOder(root.right); } } private void setInOrder(TreeNode root) { if (root != null) { setInOrder(root.left); inOrder.add(root.val); setInOrder(root.right); } } private void setPostOrder(TreeNode root) { if (root != null) { setPostOrder(root.left); setPostOrder(root.right); postOrder.add(root.val); } } }
2. 迭代遍历
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * } */ public class Solution { /** * * @param root TreeNode类 the root of binary tree * @return int整型二维数组 */ List<Integer> preOrder = new ArrayList<>(); List<Integer> inOrder = new ArrayList<>(); List<Integer> postOrder = new ArrayList<>(); public int[][] threeOrders (TreeNode root) { // write code here setPreOrder(root); setInOrder(root); setPostOrder(root); int len = preOrder.size(); int[][] ans = new int[3][len]; for (int i = 0; i < len; i++) { ans[0][i] = preOrder.get(i).intValue(); ans[1][i] = inOrder.get(i).intValue(); ans[2][i] = postOrder.get(i).intValue(); } return ans; } private void setPreOrder(TreeNode root) { if (root != null) { Stack<TreeNode> stack = new Stack<>(); TreeNode node = root; while (!stack.isEmpty() || node != null) { while (node != null) { preOrder.add(node.val); stack.push(node); node = node.left; } node = stack.pop(); node = node.right; } } } private void setInOrder(TreeNode root) { if (root != null) { Stack<TreeNode> stack = new Stack<>(); TreeNode node = root; while (!stack.isEmpty() || node != null) { while (node != null) { stack.push(node); node = node.left; } node = stack.pop(); inOrder.add(node.val); node = node.right; } } } private void setPostOrder(TreeNode root) { if (root != null) { Stack<TreeNode> stack = new Stack<>(); TreeNode prev = null; while (root != null || !stack.isEmpty()) { while (root != null) { stack.push(root); root = root.left; } root = stack.pop(); if (root.right == null || root.right == prev) { postOrder.add(root.val); prev = root; root = null; } else { stack.push(root); root = root.right; } } } } }