Leetcode - Algorithm - Binary Tree Preorder Traversal

  2017-06-03 20:36:11 +0000   |     algorithm leetcode binary tree tree stack   |   Viewed times   |    

题目

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?

递归版

Binary Tree的问题，递归法是最简单的。

代码

/**
 * Definition for a binary tree node.
 * 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> res = new ArrayList<>();
        if (root == null) { return res; }
        res.add(root.val);
        res.addAll(preorderTraversal(root.left));
        res.addAll(preorderTraversal(root.right));
        return res;
    }
}

结果

binary-tree-preorder-traversal-1

迭代版

用一个Stack缓存接下来需要处理的节点。

代码

/**
 * Definition for a binary tree node.
 * 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> res = new ArrayList<>();
        Deque<TreeNode> stack = new LinkedList<TreeNode>();
        stack.add(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pollFirst();
            if (node != null) {
                res.add(node.val);
                stack.offerFirst(node.right);
                stack.offerFirst(node.left);
            }
        }
        return res;
    }
}

结果

binary-tree-preorder-traversal-2