lintcode,二叉树的前序遍历

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给出一棵二叉树，返回其节点值的前序遍历。

解题思路：递归和非递归，三种遍历类似。

一刷ac

递归

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */public class Solution {    /**     * @param root: The root of binary tree.     * @return: Preorder in ArrayList which contains node values.     */    public ArrayList<Integer> preorderTraversal(TreeNode root) {        ArrayList<Integer> res = new ArrayList<Integer>();        if(root == null) return res;        res.add(root.val);        if(root.left != null) res.addAll(preorderTraversal(root.left));        if(root.right != null) res.addAll(preorderTraversal(root.right));        return res;    }}

非递归

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */public class Solution {    /**     * @param root: The root of binary tree.     * @return: Preorder in ArrayList which contains node values.     */    public ArrayList<Integer> preorderTraversal(TreeNode root) {        ArrayList<Integer> res = new ArrayList<Integer>();        if(root == null) return res;        Stack<TreeNode> stack = new Stack<TreeNode>();        stack.push(root);        while(!stack.empty()){            TreeNode 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;    }}

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