94. Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes' values.
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
if (root == null) {
return res;
}
TreeNode curNode = root, preNode = null;
while (curNode != null) {
if (curNode.left == null) {
res.add(curNode.val);
curNode = curNode.right;
}
else {
preNode = curNode.left;
while (preNode.right != null && preNode.right != curNode) {
preNode = preNode.right;
}
if (preNode.right == null) {
preNode.right = curNode;
curNode = curNode.left;
}
else {
res.add(curNode.val);
preNode.right = null;
curNode = curNode.right;
}
}
}
return res;
}
}
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<>();
// Method 1: Iterative Stack
TreeNode curNode = root;
Stack<TreeNode> stack = new Stack<>();
while (curNode != null || !stack.isEmpty()) {
if (curNode != null) {
stack.push(curNode);
curNode = curNode.left;
}
else {
curNode = stack.pop();
res.add(curNode.val);
curNode = curNode.right;
}
}
return res;
}
}
3. Time & Space Complexity