给你二叉树的根节点 root ,返回它节点值的 前序 遍历。
示例 1:
输入:root = [1,null,2,3]
输出:[1,2,3]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
示例 4:
输入:root = [1,2]
输出:[1,2]
示例 5:
输入:root = [1,null,2]
输出:[1,2]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-preorder-traversal
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/**
* 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;
* }
* }
*/
//非递归先序遍历
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if(root==null){
return res;
}
// 1. 先搞一个栈
Stack<TreeNode> stack = new Stack<>();
// 2. 把根节点入栈
stack.push(root);
// 3. 循环取栈顶元素并出栈
while (!stack.isEmpty()) {
TreeNode cur = stack.pop();
// 访问该元素
res.add(cur.val);
// 把右子树入栈
if (cur.right != null) {
stack.push(cur.right);
}
// 把左子树入栈
if (cur.left != null) {
stack.push(cur.left);
}
}
return res;
}
}