Description Title: Given a binary tree in preorder to return it.
Example:
Input: [. 1, null, 2,3]
. 1
\
2
/
. 3
Output: [1,3,2]
Solution one: recursion (simpler)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
List<Integer> lis = new ArrayList<>();
public List<Integer> inorderTraversal(TreeNode root) {
IF (the root == null) return LIS; // If the node is empty, the current results returned array
inorderTraversal (root.left); // left subtree preorder
lis.add(root.val);
inorderTraversal (root.right); // right subtree to be preorder
return lis;
}
}
解法二：（迭代）--基于栈的中序遍历
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
TreeNode cur = root;
while (cur != null || !stack.isEmpty()) {
if (cur != null) {
stack.push(cur);//保留当前根节点信息
cur = cur.left;
} else {
cur = stack.pop();
list.add(cur.val);
cur = cur.right;
}
}
return list;
}
}