Binary Tree Preorder Traversal

Binary Tree Preorder Traversal 

问题：

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

Recursive solution is trivial, could you do it iteratively?

思路：

　　栈的方法 弹出栈顶的时候 先加入有节点 再加入做节点 直到栈空

我的代码：

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);
        do
        {
            TreeNode node = stack.pop();
            list.add(node.val);
            if(node.right != null)  stack.push(node.right);
            if(node.left != null)   stack.push(node.left);
        }while(!stack.isEmpty());
        return list;
    }
}

他人代码：

public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        Stack<TreeNode> stack = new Stack<TreeNode>();
        List<Integer> preorder = new ArrayList<Integer>();
        
        if (root == null) {
            return preorder;
        }
        
        stack.push(root);
        while (!stack.empty()) {
            TreeNode node = stack.pop();
            preorder.add(node.val);
            if (node.right != null) {
                stack.push(node.right);
            }
            if (node.left != null) {
                stack.push(node.left);
            }
        }
        
        return preorder;
    }
}

学习之处：

• 二叉树是有两个叉的树 二分查找树是满足左子树<节点<右子树 二叉平衡树 左子树和右子树的高度相差最大为1

if(null == head) return new ArrayList<Integer>();
List<Integer> preOrder = new ArrayList<Integer>();
Stack<TreeNode> nodeStack = new Stack<TreeNode>();
nodeStack.push(head);

while(!nodeStack.isEmpty()){
TreeNode curTopNode = nodeStack.pop();
preOrder.add(curTopNode.val);
if(null != curTopNode.right) nodeStack.push(curTopNode.right);
if(null != curTopNode.left) nodeStack.push(curTopNode.left);
}

return preOrder;