Binary Tree Preorder Traversal

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Given a binary tree, return the preorder traversal of its nodes’ values.

Note:
Recursive solution is trivial, could you do it iteratively?

描述

给定一个二叉树，返回前序遍历的值，采用非递归方式

分析

非递归方式主要使用 Stack 来模拟递归

代码

递归：

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/**
 * 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> list = new ArrayList<Integer>();

        if (root != null) {

            list.add(root.val);
            list.addAll(preorderTraversal(root.left));
            list.addAll(preorderTraversal(root.right));
        }

        return list;
    }
}

非递归：

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public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {

        List<Integer> list = new ArrayList<Integer>();

        if (root == null) {
            return list;
        }

        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);

        while (!stack.empty()) {

            TreeNode treeNode = stack.pop();
            list.add(treeNode.val);

            if (treeNode.right != null) {
                stack.push(treeNode.right);
            }

            if (treeNode.left != null) {
                stack.push(treeNode.left);
            }
        }

        return list;
    }
}