1. Binary Tree Inorder Traversal

public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {

        List<Integer> ans = new ArrayList<>();
        if(root == null) return ans;

        inorder(root, ans);
        return ans;
    }

    public void inorder(TreeNode root, List<Integer> ans) {

        if(root == null)
            return;

        inorder(root.left, ans);        
        ans.add(root.val);        
        inorder(root.right, ans);

    }
}

public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {

        List<Integer> ans = new ArrayList<>();
        if(root == null) return ans;

        Stack<TreeNode> st = new Stack<>();

        TreeNode cur = root;
        while(cur!=null || !st.isEmpty()){
            while(cur != null){
                st.push(cur);
                cur = cur.left;
            }

            cur = st.pop();
            ans.add(cur.val);
            cur = cur.right;
        }

        return ans;
    }
}

2. Binary Tree Preorder Traversal

public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {

        List<Integer> ans = new ArrayList<>();

        preorder(root, ans);
        return ans;
    }

    public void preorder(TreeNode root, List<Integer> ans) {

        if(root == null)
            return;

        ans.add(root.val);
        preorder(root.left, ans);
        preorder(root.right, ans);
    }
}

public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {

        List<Integer> ans = new ArrayList<>();
        if(root == null) return ans;

        Stack<TreeNode> st = new Stack<>();
        st.push(root);

        while(!st.isEmpty()){
            TreeNode node = st.pop();
            ans.add(node.val);

            if(node.right != null) st.push(node.right);
            if(node.left  != null) st.push(node.left);
        }

        return ans;    
    }
}

3. Binary Tree Postorder Traversal

public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {

        List<Integer> ans = new ArrayList<>();

        postorder(root, ans);
        return ans;
    }

    public void postorder(TreeNode root, List<Integer> ans) {

        if(root == null)
            return;

        postorder(root.left, ans);
        postorder(root.right,ans);

        ans.add(root.val);
    }
}

public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {

        List<Integer> ans = new ArrayList<>();
        if(root == null) return ans;

        Stack<TreeNode> st = new Stack<>();
        TreeNode cur = root;

        while(cur != null || !st.isEmpty()){

            while(cur != null){
                st.push(cur);
                if(cur.left != null)
                    cur = cur.left;
                else
                    cur = cur.right;
            }

            cur = st.pop();
            ans.add(cur.val);
            if(!st.isEmpty() && st.peek().left == cur)
                cur = st.peek().right;
            else
                cur = null;
        }

        return ans;
    }
}

4. Binary Tree Zigzag Level Order Traversal

Given a binary tree, return thezigzag level ordertraversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree[3,9,20,null,null,15,7],

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

public class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {

        ArrayList<List<Integer>> ans = new ArrayList<>();
        if(root == null) return ans;

        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        boolean flag = true;

        while(!q.isEmpty()){

            List<Integer> sub = new ArrayList<>();
            int size = q.size();
            for(int i = 0; i < size; i ++){
                TreeNode node = q.poll();
                if(node.left != null) q.offer(node.left);
                if(node.right!= null) q.offer(node.right);
                if(flag)
                    sub.add(node.val);
                else
                    sub.add(0, node.val);
            }
            flag = !flag;
            ans.add(sub);
        }

        return ans;
    }
}

5. Binary Tree Level Order Traversal II

Given a binary tree, return thebottom-up level ordertraversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree[3,9,20,null,null,15,7],

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

public class Solution {

    public List<List<Integer>> levelOrderBottom(TreeNode root) {

        ArrayList<List<Integer>> ans = new ArrayList<>();
        if(root == null) return ans;

        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);

        while(!q.isEmpty()){
            int size = q.size();
            List<Integer> sub = new ArrayList<>();

            for(int i = 0; i < size; i ++){
                TreeNode node = q.poll();
                if(node.left != null) q.offer(node.left);
                if(node.right!= null) q.offer(node.right);
                sub.add(node.val);
            }
            ans.add(0, sub);
        }

        return ans;
    }
}

6. Flatten Binary Tree to Linked List

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

The flattened tree should look like:

public class Solution {

    TreeNode start = new TreeNode(0);
    TreeNode pre = start;

    public void flatten(TreeNode root) {

        if(root == null) return;

        preOrder(root);
        root = start.right;
    }

    public void preOrder(TreeNode root) {

        if(root == null)
            return;

        TreeNode left_tree = root.left;
        TreeNode right_tree= root.right;

        root.left = null;
        root.right= null;

        pre.right = root;
        pre = root;

        preOrder(left_tree);
        preOrder(right_tree);
    }
}

3.2 Binary Tree Traversal