Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes’ values.
Example :
Given binary tree
1
\
2
/
3
return [1,3,2].
Using recursion is not allowed.
Solution:
Time: O(n)
Space: O(n)
/**
* Definition for binary tree
* class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) {
* val = x;
* left=null;
* right=null;
* }
* }
*/
public class Solution {
public ArrayList<Integer> inorderTraversal(TreeNode A) {
ArrayList<Integer> result = new ArrayList<>();
Deque<TreeNode> stack = new ArrayDeque<>();
TreeNode curr = A;
while (curr != null || !stack.isEmpty()) {
while (curr != null) {
stack.push(curr);
curr = curr.left;
}
if (!stack.isEmpty()) {
curr = stack.pop();
result.add(curr.val);
curr = curr.right;
}
}
return result;
}
}