目录
:给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。
遍历(Traversal)是指沿着某条搜索路线，依次对树中每个结点均做一次且仅做一次访问 前序遍历(Preorder Traversal 亦称先序遍历)——访问根结点--->根的左子树--->根的右子树。
中序遍历(Inorder Traversal)——根的左子树--->根节点--->根的右子树。
后序遍历(Postorder Traversal)——根的左子树--->根的右子树--->根节点
二叉树的前序遍历
题目
:给你二叉树的根节点 root ，返回它节点值的 前序 遍历
前序遍历题目链接
: https://leetcode.cn/problems/binary-tree-preorder-traversal/
递归代码
1.利用方法返回值的代码
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
if(root==null){
return ret;
}
ret.add(root.val);
List<Integer> leftTree=preorderTraversal(root.left);
ret.addAll(leftTree);
List<Integer> rightTree=preorderTraversal(root.right);
ret.addAll(rightTree);
return ret;
}
}2.返回值为void的代码
// 前序遍历 根 左子树 右子树
public void preOrder(BTNode root) {
if (root == null) {
return;
}
System.out.print(root.value + " ");
preOrder(root.left);
preOrder(root.right);
}非递归实现前序遍历(利用栈stack)
1.利用方法返回值的代码
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<>();
if (root == null) {
return ret;
}
TreeNode cur = root;
Deque<TreeNode> stack = new ArrayDeque<>();
while (cur != null || !stack.isEmpty()) {
while (cur != null) {
stack.push(cur);
//System.out.print(cur.val + " ");
ret.add(cur.val);
cur = cur.left;
}
TreeNode top = stack.pop();
cur = top.right;
}
return ret;
}
}2.返回值为void的代码
class Solution {
public void preOrderNor(BTNode root) {
if (root == null) {
return;
}
BTNode cur = root;
Deque<BTNode> stack = new ArrayDeque<>();
while (cur != null || !stack.isEmpty()) {
while (cur != null) {
stack.push(cur);
System.out.println(cur.value + " ");
cur = cur.left;
}
BTNode top = stack.pop();
cur = top.right;
}
}
}二叉树的中序遍历
题目
:给定一个二叉树的根节点 root ，返回 它的 中序 遍历 。
中序遍历题目链接
:https://leetcode.cn/problems/binary-tree-inorder-traversal/
递归代码
1.利用方法返回值的代码
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<>();
if (root == null) {
return ret;
}
List<Integer> leftTree = inorderTraversal(root.left);
ret.addAll(leftTree);
ret.add(root.val);
List<Integer> rightTree =inorderTraversal(root.right);
ret.addAll(rightTree);
return ret;
}
}2.返回值为void的代码
// 中序遍历 左子树 根 右子树
public void inOrder(BTNode root) {
if (root == null) {
return;
}
inOrder(root.left);
System.out.print(root.value + " ");
inOrder(root.right);
}非递归实现中序遍历(利用栈stack)
1.利用方法返回值的代码
class Solution{
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<>();
if (root == null) {
return ret;
}
TreeNode cur = root;
Deque<TreeNode> stack = new ArrayDeque<>();
while (cur != null || !stack.isEmpty()) {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
TreeNode top = stack.pop();
ret.add(top.val);
cur = top.right;
}
return ret;
}
}2.返回值为void的代码
class Solution{
public void inOrderNor(BTNode root) {
if (root == null) {
return;
}
BTNode cur = root;
Deque<BTNode> stack = new ArrayDeque<>();
while (cur != null || !stack.isEmpty()) {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
BTNode top = stack.pop();
System.out.println(top.value + " ");
cur = top.right;
}
}
}二叉树的后序遍历
题目
:给你一棵二叉树的根节点 root ，返回其节点值的 后序遍历 。
后序遍历题目链接
:https://leetcode.cn/problems/binary-tree-postorder-traversal/
递归代码
1.利用方法返回值的代码
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
if(root==null){
return ret;
}
List<Integer> leftTree = postorderTraversal(root.left);
ret.addAll(leftTree);
List<Integer> rightTree = postorderTraversal(root.right);
ret.addAll(rightTree);
ret.add(root.val);
return ret;
}
}2.返回值为void的代码
// 后序遍历
public void postOrder(BTNode root) {
if (root == null) {
return;
}
postOrder(root.left);
postOrder(root.right);
System.out.print(root.value + " ");
}非递归实现中序遍历(利用栈stack)
1.利用方法返回值的代码
class Solution{
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> ret = new ArrayList<>();
if (root == null) {
return ret;
}
TreeNode cur = root;
TreeNode prev = null;
Deque<TreeNode> stack = new ArrayDeque<>();
while (cur != null || !stack.isEmpty()) {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
TreeNode top = stack.peek();
if (top.right == null || top.right == prev) {
//System.out.println(top.value + " ");
ret.add(top.val);
stack.pop();
prev = top;
} else {
cur = top.right;
}
}
return ret;
}
} 2.返回值为void的代码
class Solution{
public void postOrderNor(BTNode root) {
if (root == null) {
return;
}
BTNode cur = root;
BTNode prev = null;
Deque<BTNode> stack = new ArrayDeque<>();
while (cur != null || !stack.isEmpty()) {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
BTNode top = stack.peek();
if (top.right == null || top.right == prev) {
System.out.println(top.value + " ");
stack.pop();
prev = top;
} else {
cur = top.right;
}
}
}
}完结撒花✿✿ヽ(°▽°)ノ✿✿
