Binary Tree Preorder Traversal

Given a binary tree, return the _preorder_ traversal of its nodes' values.
Example:
Input:
[1,null,2,3]
1
\
2
/
3
Output:
[1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?

Solution

Recursive

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
recursivePreorderTraversal(root, result);
return result;
}
void recursivePreorderTraversal(TreeNode root, List<Integer> result) {
if (root != null) {
result.add(root.val);
recursivePreorderTraversal(root.left, result);
recursivePreorderTraversal(root.right, result);
}
}
}

*LeetCode Official Solution* - Iterative using Stack

注意node.right, node.left加入stack的顺序,因为stack是FILO,所以先放入right子节点,再放left子节点
/* Definition for a binary tree node. */
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
LinkedList<TreeNode> stack = new LinkedList<>();
LinkedList<Integer> output = new LinkedList<>();
if (root == null) {
return output;
}
stack.add(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pollLast();
output.add(node.val);
if (node.right != null) {
stack.add(node.right);
}
if (node.left != null) {
stack.add(node.left);
}
}
return output;
}
}

Iterative - Using Stack

public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<>();
Deque<TreeNode> stack = new ArrayDeque<>();
TreeNode p = root;
while(!stack.isEmpty() || p != null) {
if(p != null) {
stack.push(p);
result.add(p.val); // Add before going to children
p = p.left;
} else {
TreeNode node = stack.pop();
p = node.right;
}
}
return result;
}

Another Stack implementation

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
Stack <TreeNode> stack = new Stack <TreeNode> ();
TreeNode curr = root;
while (curr != null || !stack.isEmpty()) {
while (curr != null) {
result.add(curr.val);
stack.push(curr);
curr = curr.left;
}
curr = stack.pop();
curr = curr.right;
}
return result;
}
}

Morris Traversal

class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
LinkedList<Integer> output = new LinkedList<>();
TreeNode node = root;
while (node != null) {
if (node.left == null) {
output.add(node.val);
node = node.right;
}
else {
TreeNode predecessor = node.left;
while ((predecessor.right != null) && (predecessor.right != node)) {
predecessor = predecessor.right;
}
if (predecessor.right == null) {
output.add(node.val);
predecessor.right = node;
node = node.left;
}
else{
predecessor.right = null;
node = node.right;
}
}
}
return output;
}
}

Reference