Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,2,3].
Note: Recursive solution is trivial, could you do it iteratively?
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<Integer> preorderTraversal(TreeNode root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
ArrayList<Integer> res = new ArrayList<Integer>();
if(root == null)
return res;
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
TreeNode node;
while(!stack.isEmpty()){
node = stack.pop();
res.add(node.val);
if(node.right != null)
stack.push(node.right);
if(node.left != null)
stack.push(node.left);
}
return res;
}
}

本文介绍了一种不使用递归的方法来完成二叉树节点值的前序遍历,通过栈来实现对节点的访问顺序,提供了一个具体的Java实现案例。
122

被折叠的 条评论
为什么被折叠?



