前序遍历
题目
给定一个二叉树,返回它的 前序 遍历。
示例:
输入: [1,null,2,3]
1
2
/
3
输出: [1,2,3]
思路
迭代
从根节点开始,每次迭代弹出当前栈顶元素,并将其孩子节点压入栈中,先压右孩子再压左孩子。
代码
package com.alone.test.leetcode;
import javax.naming.LinkLoopException;
import java.util.LinkedList;
import java.util.List;
/**
* @BelongsProject: JavaSourceNotes
* @BelongsPackage: com.alone.test.leetcode
* @Author: Alone
* @CreateTime: 2020-08-20 07:37
* @Description: 二叉树的前序遍历
*/
public class T144 {
private List<Integer> result = new LinkedList<>();
public List<Integer> preorderTraversal(TreeNode root) {
preOrderStack(root);
return result;
}
/**
* 递归
* @param root
*/
private void preOrder(TreeNode root) {
if(root == null)
return;
result.add(root.val);
preOrder(root.left);
preOrder(root.right);
}
/**
* 栈迭代
* @param root
*/
private void preOrderStack(TreeNode root) {
LinkedList<TreeNode> stack = new LinkedList<>();
if(root != null)
stack.add(root);
while (!stack.isEmpty()) {
TreeNode now = stack.pollLast();//获取的同时删除
result.add(now.val);
//要反过来,因为先遍历左子树
if(now.right != null)
stack.add(now.right);
if(now.left != null)
stack.add(now.left);
}
}
/**
* 莫里斯算法 时间复杂度O(n) 空间复杂度O(1)
* 有点不太理解
* @param root
*/
private void preOrderMorris(TreeNode root) {
TreeNode node = root;
while (node != null) {
if (node.left == null) {//左子树为空
result.add(node.val);
node = node.right;
}
else {
TreeNode predecessor = node.left;
while ((predecessor.right != null) && (predecessor.right != node)) {//找到右子树的叶节点
predecessor = predecessor.right;
}
if (predecessor.right == null) {//把当前节点与根节点连接
result.add(node.val);
predecessor.right = node;
node = node.left;
}
else{//删除虚拟路径
predecessor.right = null;
node = node.right;
}
}
}
}
}
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
中序遍历
思路
迭代
思路:每到一个节点 A,因为根的访问在中间,将 A 入栈。然后遍历左子树,接着访问 A,最后遍历右子树。
在访问完 A 后,A 就可以出栈了。因为 A 和其左子树都已经访问完成。
代码
package com.alone.test.leetcode;
import java.util.LinkedList;
import java.util.List;
/**
* @BelongsProject: JavaSourceNotes
* @BelongsPackage: com.alone.test.leetcode
* @Author: Alone
* @CreateTime: 2020-08-20 09:55
* @Description: 二叉树的中序遍历
*/
public class T94 {
private List<Integer> result = new LinkedList<>();
public List<Integer> inorderTraversal(TreeNode root) {
inOrderStack(root);
return result;
}
/**
* 递归
* @param root
*/
private void inOrder(TreeNode root) {
if(root == null)
return;
inOrder(root.left);
result.add(root.val);
inOrder(root.right);
}
/**
* 迭代
* @param root
*/
private void inOrderStack(TreeNode root) {
LinkedList<TreeNode> stack = new LinkedList<>();
if(root != null){
stack.add(root);
}
TreeNode tmp = root;
while (tmp != null && !stack.isEmpty()) {
while(tmp != null){
stack.add(tmp);
tmp = tmp.left;
}
tmp = stack.pollLast();
result.add(tmp.val);
tmp = tmp.right;
}
}
}
后序遍历
代码
package com.alone.test.leetcode;
import java.util.LinkedList;
import java.util.List;
/**
* @BelongsProject: JavaSourceNotes
* @BelongsPackage: com.alone.test.leetcode
* @Author: Alone
* @CreateTime: 2020-08-22 09:35
* @Description:
*/
public class T145 {
private LinkedList<Integer> result = new LinkedList<>();
public List<Integer> postorderTraversal(TreeNode root) {
postOrderStack(root);
return result;
}
private void postOrder(TreeNode root) {
if(root == null)
return;
postOrder(root.left);
postOrder(root.right);
result.add(root.val);
}
private void postOrderStack(TreeNode root) {
LinkedList<TreeNode> stack = new LinkedList<>();
if(root != null)
stack.add(root);
while (!stack.isEmpty()) {
TreeNode tmp = stack.pollLast();
result.addFirst(tmp.val);
if(tmp.left != null)
stack.add(tmp.left);
if(tmp.right != null)
stack.add(tmp.right);
}
}
}
层序遍历
题目
给你一个二叉树,请你返回其按 层序遍历 得到的节点值。 (即逐层地,从左到右访问所有节点)。
示例:
二叉树:[3,9,20,null,null,15,7],
3
/
9 20
/
15 7
返回其层次遍历结果:
[
[3],
[9,20],
[15,7]
]
思路
采用bfs,先统计每一层的节点数,再对每一层节点进行遍历。
代码
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.Queue;
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
Queue<TreeNode> queue = new ArrayDeque<>();
if (root == null)
return result;
else {
queue.add(root);
while (!queue.isEmpty()) {
int s = queue.size();
List<Integer> tmp = new ArrayList<>();
while (s > 0) {
TreeNode node = queue.poll();
tmp.add(node.val);
if(node.left != null)
queue.add(node.left);
if(node.right != null)
queue.add(node.right);
s --;
}
if(tmp != null)
result.add(tmp);
}
}
return result;
}
}