Leetcode-144：二叉树前序遍历

递归方法：

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

非递归方法：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root==null)
            return res;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode p = root;
        //边遍历边打印，并存入栈中，以后需要借助这些节点进入右子树
        while(!stack.isEmpty() || p!=null ){
            while(p!=null){
                res.add(p.val); //先访问，与中序不同
                stack.push(p);
                p=p.left;
            }
            if(!stack.isEmpty()){
                p = stack.pop();
                p=p.right;
            }         
        }
        return res;
    }
}

非递归第二种方法：（我个人更喜欢这种写法，比较好想，方便记忆）
因为前序是根左右的访问顺序，所以利用一个栈，从根节点开始，访问该节点，然后按照右左子树节点的顺序入栈，注意是右左！因为用的是栈，先进后出，这样下一个访问的节点就是左子树节点啦~

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