144. Binary Tree Preorder Traversal (M)

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?

题意

求树的前序遍历。

思路

递归或者迭代。

代码实现 - 递归

/**
 * 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> ans = new ArrayList<>();
        preOrder(root, ans);
        return ans;
    }
    private void preOrder(TreeNode x, List<Integer> ans) {
        if (x == null) {
            return;
        }
        ans.add(x.val);
        preOrder(x.left, ans);
        preOrder(x.right, ans);
    }
}

代码实现 - 迭代

/**
 * 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> ans = new ArrayList<>();
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode cur = root;
        while (cur != null || !stack.isEmpty()) {
            if (cur != null) {
                ans.add(cur.val);
                stack.push(cur);
                cur = cur.left;
            } else {
                cur = stack.pop().right;
            }
        }
        return ans;
    }
}