Leetcode 144 Binary Tree Preorder Traversal

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

Example 1:

1 2	Input: root = [1,null,2,3] Output: [1,2,3]

Example 2:

1 2	Input: root = [] Output: []

Example 3:

1 2	Input: root = [1] Output: [1]

Example 4:

1 2	Input: root = [1,2] Output: [1,2]

Example 5:

1 2	Input: root = [1,null,2] Output: [1,2]

Constraints:

The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 100

Follow up:

Recursive solution is trivial, could you do it iteratively?

非递归

参考于 https://www.cnblogs.com/dolphin0520/archive/2011/08/25/2153720.html

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> preorder = new ArrayList<>();
        Stack<TreeNode> stk = new Stack<>();
        TreeNode p = root;
        while(p != null || !stk.isEmpty())
        {
            while(p != null)
            {
                preorder.add(p.val);
                stk.push(p);
                p = p.left;
            }
            p = stk.pop();
            p = p.right;
        }
        return preorder;
    }
}