[leetcode] 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?

---

分析：也就是求二叉树的先序遍历。

思路一：递归，DFS求解。比较简单

class Solution {
    List<Integer> res = new ArrayList<Integer>();
    public List<Integer> preorderTraversal(TreeNode root) {
        helper(root);
        return res;
    }
    private void helper(TreeNode root){
        if ( root == null ) return ;
        res.add(root.val);
        helper(root.left);
        helper(root.right);
    }
}

思路二：非递归方法实现。真正面试如果考到二叉树的三种遍历方法，肯定是考非递归，也就是栈的方法来实现的。

参考：https://www.cnblogs.com/boris1221/p/9398848.html

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        if ( root == null ) return list;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while ( !stack.isEmpty() ){
            while ( root != null ){
                list.add(root.val);
                stack.push(root);
                root = root.left;
            }
            if ( !stack.isEmpty() ){
                root = stack.pop().right;
            }
        }
        return list;
    }
}