二叉树的前序遍历

题目

前序遍历题目链接

递归代码

1.利用方法返回值的代码

2.返回值为void的代码

非递归实现前序遍历(利用栈stack)

1.利用方法返回值的代码

2.返回值为void的代码

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

递归代码

1.利用方法返回值的代码

2.返回值为void的代码

非递归实现中序遍历(利用栈stack)

1.利用方法返回值的代码

2.返回值为void的代码

二叉树的后序遍历

题目

后序遍历题目链接

递归代码

1.利用方法返回值的代码

2.返回值为void的代码

非递归实现中序遍历(利用栈stack)

1.利用方法返回值的代码

2.返回值为void的代码

完结撒花✿✿ヽ(°▽°)ノ✿✿

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;}}}
}