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# Binary Tree Preorder Traversal

Given a binary tree, return the \_preorder\_ traversal of its nodes' values.

**Example:**

```
Input:
[1,null,2,3]

   1
    \
     2
    /
   3


Output:
[1,2,3]
```

**Follow up:** Recursive solution is trivial, could you do it iteratively?

## Solution

### **Recursive**

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        recursivePreorderTraversal(root, result);
        return result;
    }
    void recursivePreorderTraversal(TreeNode root, List<Integer> result) {
        if (root != null) {
            result.add(root.val);
            recursivePreorderTraversal(root.left, result);
            recursivePreorderTraversal(root.right, result);
        }
    }
}
```

### \***LeetCode Official Solution\* - Iterative using Stack**

注意node.right, node.left加入stack的顺序，因为stack是FILO，所以先放入right子节点，再放left子节点

```java
/* Definition for a binary tree node. */
public class TreeNode {
  int val;
  TreeNode left;
  TreeNode right;

  TreeNode(int x) {
    val = x;
  }
}

class Solution {
  public List<Integer> preorderTraversal(TreeNode root) {
    LinkedList<TreeNode> stack = new LinkedList<>();
    LinkedList<Integer> output = new LinkedList<>();
    if (root == null) {
      return output;
    }

    stack.add(root);
    while (!stack.isEmpty()) {
      TreeNode node = stack.pollLast();
      output.add(node.val);
      if (node.right != null) {
        stack.add(node.right);
      }
      if (node.left != null) {
        stack.add(node.left);
      }
    }
    return output;
  }
}
```

### Iterative - Using Stack

```java
public List<Integer> preorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<>();
    Deque<TreeNode> stack = new ArrayDeque<>();
    TreeNode p = root;
    while(!stack.isEmpty() || p != null) {
        if(p != null) {
            stack.push(p);
            result.add(p.val);  // Add before going to children
            p = p.left;
        } else {
            TreeNode node = stack.pop();
            p = node.right;   
        }
    }
    return result;
}
```

### Another Stack implementation

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        Stack <TreeNode> stack = new Stack <TreeNode> ();
        TreeNode curr = root;
        while (curr != null || !stack.isEmpty()) {
            while (curr != null) {
                result.add(curr.val);
                stack.push(curr);
                curr = curr.left;
            }
            curr = stack.pop();
            curr = curr.right;
        }
        return result;
    }

}
```

### Morris Traversal

```java
class Solution {
  public List<Integer> preorderTraversal(TreeNode root) {
    LinkedList<Integer> output = new LinkedList<>();

    TreeNode node = root;
    while (node != null) {
      if (node.left == null) {
        output.add(node.val);
        node = node.right;
      }
      else {
        TreeNode predecessor = node.left;
        while ((predecessor.right != null) && (predecessor.right != node)) {
          predecessor = predecessor.right;
        }

        if (predecessor.right == null) {
          output.add(node.val);
          predecessor.right = node;
          node = node.left;
        }
        else{
          predecessor.right = null;
          node = node.right;
        }
      }
    }
    return output;
  }
}
```

## Reference

<https://leetcode.com/problems/binary-tree-preorder-traversal/solution/>


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