package gitLeetCode;
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;
/**
* 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].
*
* @author chenshuna
*
* Note: Recursive solution
*/
public class BinaryTreePreorderTraversal {
/**
* Recursive solution
*/
public static List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null) {
return res;
}
preorderTraversalTree(root, res);
return res;
}
public static void preorderTraversalTree(TreeNode root, List<Integer> res) {
if (root != null) {
res.add(root.val);
preorderTraversalTree(root.left, res);
preorderTraversalTree(root.right, res);
} else return;
}
/**
* iterative solution
* Use stack
*/
public static List<Integer> preorderTraversalIterative(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null) {
return res;
}
Stack<TreeNode> stack = new Stack<TreeNode>();
stack.push(root);
while (!stack.isEmpty()) {
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;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode res = new TreeNode(0);
res.left = new TreeNode(2);
res.right = new TreeNode(1);
res.left.left = new TreeNode(4);
res.left.right = new TreeNode(5);
res.left.left.left = new TreeNode(9);
System.out.print(preorderTraversal(res));
System.out.print(preorderTraversalIterative(res));
}
}