层序遍历——队列实现

给你二叉树的根节点 root ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）

分析

借助队列存储的方式实现。队列这个数据结构是先入先出的。

具体步骤：
1、将根节点入队
2、出队首节点，将队首节点的左右非空孩子入队
3、重复2操作直到队列为空

注意：Java的队列由linkedList实现的，这个需要注意一下。

Java完整代码

/*** 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>> levelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<>();  if (root == null) {return res;}LinkedList<TreeNode> queue = new LinkedList<>(); // Java的队列由linkedList实现的queue.add(root);while (!queue.isEmpty()) {List<Integer> list = new ArrayList<>();int current_queue_size = queue.size();for (int i = 0; i < current_queue_size; i++) {TreeNode top = queue.getFirst();list.add(top.val);if (top.left != null) {queue.add(top.left);}if (top.right != null) {queue.add(top.right);}queue.removeFirst();}res.add(list);}return  res;}}

先序遍历——中左右

给你二叉树的根节点 root ，返回它节点值的 前序 遍历。

分析

前序遍历是先访问根节点再左孩子然后右孩子。

递归实现

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);}
}

非递归实现——栈实现

借助栈数据结构

 * 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) {List<Integer> result = new ArrayList<>();if (root == null) {return result;}Stack<TreeNode> stack = new Stack<>();TreeNode p = root;TreeNode tmp;while (p != null || !stack.isEmpty()) {// 一路向左if (p != null) {result.add(p.val);stack.push(p);p = p.left;} else {tmp = stack.pop();p = tmp.right;}}return result;}
}

中序遍历——左中右

递归实现

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);}
}

非递归实现——栈实现

/*** 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> result = new ArrayList<>();if (root == null) {return result;}Stack<TreeNode> stack = new Stack<>();TreeNode p = root;TreeNode tmp;while (p != null || !stack.isEmpty()) {// 一路向左if (p != null) {stack.push(p);p = p.left;} else {tmp = stack.pop();result.add(tmp.val);p = tmp.right;}}return result;}
}

后续遍历——左右中

递归实现

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);}
}

非递归实现——栈加标志指针实现

/*** 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) {List<Integer> result = new ArrayList<>();if (root == null) {return result;}Stack<TreeNode> stack = new Stack<>();TreeNode p = root;TreeNode tmp;TreeNode review = null;while (p != null || !stack.isEmpty()) {// 一路向左if (p != null) {stack.push(p);p = p.left;} else {tmp = stack.peek();if (tmp.right == null || review == tmp.right) {result.add(tmp.val);review = stack.pop();} else {p = tmp.right;}}}return result;}
}

总结

对于树的问题，更多能通过递归去简化问题，更好解决实际问题。

ps:计划每日更新一篇博客，今日2023-04-29，日更第十三天。
昨日更新：
反转字符串