144 Binary Tree Preorder Traversal

144. Binary Tree Preorder Traversal

1. Question

Given a binary tree, return thepreordertraversal of its nodes' values.

For example: Given binary tree{1,#,2,3},

   1
    \
     2
    /
   3

return[1,2,3].

2. Implementation

(1) Morris Tree Traversal

class Solution {
    public List<Integer> preorderTraversal(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) {
                    res.add(curNode.val);
                    preNode.right = curNode;
                    curNode = curNode.left;
                }
                else {
                    preNode.right = null;
                    curNode = curNode.right;
                }
            }
        }
        return res;
    }
}

(2) Iteration

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();

        if (root == null) {
            return res;
        }

        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        TreeNode curNode = null;

        while (!stack.isEmpty()) {
            curNode = stack.pop();
            res.add(curNode.val);

            if (curNode.right != null) {
                stack.push(curNode.right);
            }

            if (curNode.left != null) {
                stack.push(curNode.left);
            }
        }
        return res;
    }
}

3. Time & Space Complexity

(1) Morris Tree Traversal: 时间复杂度: O(n), 空间复杂度: O(1)

(2) Iteration: 时间复杂度: O(n), 空间复杂度: O(h)