题目描述
思路
熟练掌握迭代和递归的代码。
- 递归代码：额外写一个函数void preOrder(TreeNode node, List
res) - 迭代代码：会用到数据结构——栈。先入栈当前节点的右子节点，再入栈左子节点。
方法一：递归
/**
* 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> res = new ArrayList<>();
preOrder(root, res);
return res;
}
private void preOrder(TreeNode node, List<Integer> res) {
if (node == null) return;
res.add(node.val);
preOrder(node.left, res);
preOrder(node.right, res);
}
}
方法二：迭代
/**
* 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> res = new ArrayList<>();
Deque<TreeNode> stack = new ArrayDeque<>();
if (root != null) stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
res.add(node.val);
// 先入栈当前节点的右子节点，再入栈左子节点。
if (node.right != null) stack.push(node.right);
if (node.left != null) stack.push(node.left);
}
return res;
}
}