文章目录
- 二叉树递归遍历
- 解题思路
- 代码
- 总结
- 二叉树的迭代遍历
- 解题思路
- 代码
- 总结
- 二叉树的统一迭代法
- 解题思路
- 代码
- 总结
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java的Deque
二叉树递归遍历
题目：
144.二叉树的前序遍历
94.二叉树的中序遍历
145.二叉树的后序遍历
解析：代码随想录解析
解题思路
递归遍历
前序：NLR
中序：LNR
后序：LRN
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;
* }
* }
*/
//前序
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
preorder(root, res);
return res;
}
public void preorder(TreeNode root, List<Integer> res){
if (root == null)
return;
res.add(root.val);
preorder(root.left, res);
preorder(root.right, res);
}
}
//中序
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
inorder(root, res);
return res;
}
public void inorder(TreeNode root, List<Integer> res){
if (root == null)
return;
inorder(root.left, res);
res.add(root.val);
inorder(root.right, res);
}
}
//后序
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
postorder(root, res);
return res;
}
public void postorder(TreeNode root, List<Integer> res){
if (root == null)
return;
postorder(root.left, res);
postorder(root.right, res);
res.add(root.val);
}
}
总结
暂无
二叉树的迭代遍历
题目：
144.二叉树的前序遍历
94.二叉树的中序遍历
145.二叉树的后序遍历
解析：代码随想录解析
解题思路
前序：利用一个栈，每次出栈并入栈。
中序：利用一个栈，cur指向root节点，一直走左子树并入栈到空；cur为空时输出栈顶的val，然后使cur指向出栈节点右子树，重复上述步骤。
后序：LRN反过来是NRL，也就是前序换一下，最后倒转一下。
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;
* }
* }
*/
//前序
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null)
return res;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode tmp = stack.pop();
res.add(tmp.val);
if (tmp.right != null)
stack.push(tmp.right);
if (tmp.left != null)
stack.push(tmp.left);
}
return res;
}
}
//中序
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null)
return res;
Stack<TreeNode> stack = new Stack<TreeNode>();
TreeNode cur = root;
while (!stack.isEmpty() || cur != null){
if (cur != null){
stack.push(cur);
cur = cur.left;
}else{
cur = stack.pop();
res.add(cur.val);
cur = cur.right;
}
}
return res;
}
}
//后序
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null)
return res;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode tmp = stack.pop();
res.add(tmp.val);
if (tmp.left != null)
stack.push(tmp.left);
if (tmp.right != null)
stack.push(tmp.right);
}
Collections.reverse(res);
return res;
}
}
总结
死去的408记忆在攻击我
二叉树的统一迭代法
题目：
144.二叉树的前序遍历
94.二叉树的中序遍历
145.二叉树的后序遍历
解析：代码随想录解析
解题思路
代码结构和递归遍历相似。下面是模拟步骤图
前序
中序
后序：
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;
* }
* }
*/
//前序
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null)
return res;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode node = stack.peek();
if (node != null){
stack.pop();
if (node.right != null) stack.push(node.right);
if (node.left != null) stack.push(node.left);
stack.push(node);
stack.push(null);
}else{
stack.pop();
node = stack.pop();
res.add(node.val);
}
}
return res;
}
}
//中序
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null)
return res;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode node = stack.peek();
if (node != null){
stack.pop();
if (node.right != null) stack.push(node.right);
stack.push(node);
stack.push(null);
if (node.left != null) stack.push(node.left);
}else{
stack.pop();
node = stack.pop();
res.add(node.val);
}
}
return res;
}
}
//后序
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null)
return res;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while(!stack.isEmpty()){
TreeNode node = stack.peek();
if (node != null){
stack.pop();
stack.push(node);
stack.push(null);
if (node.right != null) stack.push(node.right);
if (node.left != null) stack.push(node.left);
}else{
stack.pop();
node = stack.pop();
res.add(node.val);
}
}
return res;
}
}
总结
感觉记住了，感觉。
