Leetcode 144. Preorder traversal of binary tree
Title description
Give you the root node root of the binary tree, and return the preorder traversal of its node value.
Example 1:
输入：root = [1,null,2,3]
输出：[1,2,3]
Example 2:
输入：root = []
输出：[]
Example 3:
输入：root = [1]
输出：[1]
Example 4:
输入：root = [1,2]
输出：[1,2]
Example 5:
输入：root = [1,null,2]
输出：[1,2]
prompt:
The number of nodes in the tree is within the range [0, 100]
-100 <= Node.val <= 100
Advanced: The recursive algorithm is very simple, can you do it through an iterative algorithm?
Place.
Solution 1: Recursion
Design a function and then recursively call to achieve traversal
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> list = new LinkedList<>();
preorderTraversal(root,list);
return list;
}
public void preorderTraversal(TreeNode root, List<Integer> list) {
if (root==null){
return;
}
list.add(root.val);
preorderTraversal(root.left,list);
preorderTraversal(root.right,list);
}
Solution two: non-recursive
template
The traversal of the binary tree can be thought of based on this template. On this basis, you can always think about it carefully.
while( 栈非空 || p 非空)
{
if( p 非空)
{
}
else
{
}
}
Use the stack to achieve
1. First push the root node onto the stack
2. Perform the following operations in a loop:
pop the top element of
the stack, push the left node of
the top node of the stack, push the child node of the top node of the stack,
public class Leetcode144_2 {
public List<Integer> preorderTraversal(TreeNode root) {
LinkedList<Integer> list = new LinkedList<>();
if (root==null){
return list ;
}
Stack<TreeNode> stack = new Stack<>();
TreeNode node=root;
stack.add(node);
while(!stack.isEmpty()){
node=stack.pop();
list.add(node.val);
if (node.right!=null){
stack.add(node.right);
}
if (node.left!=null){
stack.add(node.left);
}
}
return list;
}
