题目描述:
标签:栈 树
给你二叉树的根节点
root
,返回它节点值的 前序 遍历。
代码:
思路分析:前序遍历是根-左子树-右子树。思路同二叉树的中序遍历
《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> list = new ArrayList<Integer>();
preorder(root,list);
return list;
}
public void preorder(TreeNode root,List<Integer> res){
if(root == null){
return;
}
res.add(root.val);
preorder(root.left,res);
preorder(root.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> list = new ArrayList<Integer>();
if(root == null){
return list;
}
Deque<TreeNode> stack = new LinkedList<TreeNode>();
while(root != null || !stack.isEmpty()){
while(root != null){
list.add(root.val);
stack.push(root);
root = root.left;
}
root = stack.pop();
root = root.right;
}
return list;
}
}