LeetCode 144 Binary Tree Preorder Traversal (先序遍历二叉树)

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

题目链接：https://leetcode.com/problems/binary-tree-preorder-traversal/

题目分析：

递归

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

非递归的方法就是拿栈模拟，中往左走的时候直接遍历，走到不能走然后再往右

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        while(root != null || stack.size() > 0) {
            while(root != null) {
                ans.add(root.val);
                stack.push(root);
                root = root.left;
            }
            if(stack.size() > 0) {
                root = stack.peek();
                stack.pop();
                root = root.right;
            }
        }
        return ans;
    }
}