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LeetCode-Binary Tree Preorder Traversal

时间:2015-03-12 12:48:57      阅读:95      评论:0      收藏:0      [点我收藏+]

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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?

这道题考察用 stack实现前序遍历,代码如下:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> list = new ArrayList<Integer>();
        Stack<TreeNode> stack =new Stack<TreeNode>();
        if(root == null){
            return list;
        }
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            list.add(node.val);
            if(node.right != null){
                stack.push(node.right);
            }
            if(node.left != null){
                stack.push(node.left);
            }
        }
        return list;
        
    }
}

 

LeetCode-Binary Tree Preorder Traversal

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原文地址:http://www.cnblogs.com/incrediblechangshuo/p/4331980.html

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