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# 144.Binary-Tree-Preorder-Traversal

## 144. Binary Tree Preorder Traversal

## 题目地址

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

## 题目描述

```
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?
```

## 代码

### Approach #1 Recurion

```java
class Solution {
  public List<Integer> preorderTraversal(TreeNode root) {
    List<Integer> ans = new LinkedList<Integer>();
    dfs(root, ans);
    return asn;
  }

  private void dfs(TreeNode root, List<Integer> ans) {
    if (root == null)    return;
    ans.add(root.val);
    dfs(root.left, ans);
    dfs(root.right, ans);
  }
}
```

### Approach #2 Iterations

```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<TreeNode> 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;
  }
}
```

### Approach #3 Morris Traversal Confusion

```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;
  }
}
```


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