二叉树的前、中、后序遍历
前序遍历
LC 144.二叉树的前序遍历：https://leetcode-cn.com/problems/binary-tree-preorder-traversal/
解法一：递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
List<Integer> list = new ArrayList<>();
public List<Integer> preorderTraversal(TreeNode root) { //递归实现前序遍历
if(root == null) return list;
list.add(root.val);
list = preorderTraversal(root.left);
list = preorderTraversal(root.right);
return list;
}
} 解法二：栈
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) { //使用栈实现前序遍历
List<Integer> list = new ArrayList<>();
if(root == null) return list;
Stack<TreeNode> stack = new Stack<>();
TreeNode currNode = root;
while(currNode != null || !stack.isEmpty()){
while(currNode != null){ //遍历入栈左节点，入栈的顺序就是前序遍历的顺序
list.add(currNode.val);
stack.push(currNode);
currNode = currNode.left;
}
if(!stack.isEmpty()){
currNode = stack.pop();
currNode = currNode.right; //出栈当前节点，遍历其右节点
}
}
return list;
}
} 中序遍历
解法一：递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
List<Integer> list = new ArrayList<>();
public List<Integer> inorderTraversal(TreeNode root) { //递归实现中序遍历
if(root == null) return list;
list = inorderTraversal(root.left);
list.add(root.val);
list = inorderTraversal(root.right);
return list;
}
} 解法二：栈
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
if(root == null) return list;
Stack<TreeNode> stack = new Stack<>();
TreeNode currNode = root;
while(currNode != null || !stack.isEmpty()){
while(currNode != null){ //有左节点就循环入栈左节点
stack.push(currNode);
currNode = currNode.left;
}
if(!stack.isEmpty()){
currNode = stack.pop();
list.add(currNode.val); //出栈的顺序就是中序遍历的顺序
currNode = currNode.right; //判断当前节点的右节点
}
}
return list;
}
} 后序遍历
解法一：递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
List<Integer> list = new ArrayList<>();
public List<Integer> postorderTraversal(TreeNode root) { //递归实现后序遍历
if(root == null) return list;
list = postorderTraversal(root.left);
list = postorderTraversal(root.right);
list.add(root.val);
return list;
}
} 解法二：栈
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> postorderTraversal(TreeNode root) { //使用栈实现后序遍历
List<Integer> list = new ArrayList<>();
if(root == null) return list;
Stack<TreeNode> stack = new Stack<>();
TreeNode currNode = root;
while(currNode != null || !stack.isEmpty()){
while(currNode != null){ //与前序遍历类似，但是是遍历入栈右节点，出栈遍历左节点
list.add(currNode.val);
stack.push(currNode);
currNode = currNode.right;
}
if(!stack.isEmpty()){
currNode = stack.pop().left;
}
}
Collections.reverse(list); //最后再反转List集合
return list;
}
} 二叉树的层序遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) { //使用队列实现层序遍历
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()){
int size = queue.size(); //记录当前层级的节点数
List<Integer> list = new ArrayList<>();
for(int i = 0; i < size; i++){ //遍历当前层级的所有节点
TreeNode currNode = queue.poll();
list.add(currNode.val);
if(currNode.left != null) queue.offer(currNode.left); //左右节点不为空就将其入队
if(currNode.right != null) queue.offer(currNode.right);
}
res.add(list);
}
return res;
}
} 二叉树的之字形层序遍历
NC14 按之字形顺序打印二叉树
思路
- 第一层从左向右，下一层从右向左，一直这样交替
- 思路基本与二叉树的层序遍历一样，刚开始想过利用栈或者根据不同层级决定入队的顺序，但都无法得到正确的结果
- 既然无法通过改变出入队来实现交替遍历，那就从将它加入ArrayList时入手
- 按照正常的出队，将元素加入到当前层级的ArrayList中，在偶数层时进行头插入，这样后一个元素就在前一个的前面了，在奇数层时进行尾插入
实现代码
import java.util.*;
/*
* public class TreeNode {
* int val = 0;
* TreeNode left = null;
* TreeNode right = null;
* }
*/
public class Solution {
/**
*
* @param root TreeNode类
* @return int整型ArrayList<ArrayList<>>
*/
public ArrayList<ArrayList<Integer>> zigzagLevelOrder (TreeNode root) {
// write code here
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if(root == null){
return res;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
int level = 1;//层级
while(!queue.isEmpty()){
ArrayList<Integer> arrList = new ArrayList<Integer>();
int size = queue.size(); //当前层级的元素数
for(int i = 0; i<size; i++){//将该层元素遍历出队
TreeNode node = queue.poll();
if(level % 2 == 0){//偶数层进行头插入
arrList.add(0, node.val);
}else{//奇数层进行尾插入
arrList.add(node.val);
}
if(node.left != null){
queue.offer(node.left);
}
if(node.right != null){
queue.offer(node.right);
}
}
level++;
res.add(arrList);
}
return res;
}
} 二叉树的自底向上层序遍历
LC 107. 二叉树的层序遍历 II：https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/
思路
遍历过程与正常层序遍历类似，只是将每一层的结果以头插入的方式放入结果集合中，ArrayList的add()方法正常是进行尾插入的，add(0, list)可实现头插入
实现代码
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> resList = new ArrayList<List<Integer>>();
if(root == null){
return resList;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()){
int size = queue.size(); //当前层级的节点数
List<Integer> list = new ArrayList<>();
for(int i = 0; i < size; i++){ //遍历当前层级的所有节点
TreeNode node = queue.poll();
list.add(node.val);
if(node.left != null){ //当前节点有左右节点就进行入队
queue.offer(node.left);
}
if(node.right != null){
queue.offer(node.right);
}
}
resList.add(0, list); //头插入
}
return resList;
}
} 
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