LeetCode——Binary Tree Preorder Traversal

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?

中文：二叉树的前序遍历(根-左-右)。能用非递归实现吗？

递归：

public class BinaryTreePreorderTraversal {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
if(root == null)
return list;
list.add(root.val);
list.addAll(preorderTraversal(root.left));
list.addAll(preorderTraversal(root.right));
return list;
}
// Definition for binary tree
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
}非递归：先把右节点的值压入栈中，再压入左的；弹出左的，弹出右的……。

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