LeetCode 144. Binary Tree Preorder Traversal

原题链接在这里：https://leetcode.com/problems/binary-tree-preorder-traversal/

题目：

Given a binary tree, return the preorder traversal of its nodes' values.

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

   1
    
     2
    /
   3 

return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

题解：

类似Binary Tree Inorder Traversal, Binary Tree Postorder Traversal. 

Method 1 是Recursion.

Time Complexity: O(n), 每个点访问了一遍. Space: O(logn), stack space.

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    // Method 1: Recursion
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ls = new ArrayList<Integer>();
        preorderTraversal(root,ls);
        return ls;
        
    }
    public void preorderTraversal(TreeNode root, List<Integer> ls) {
        if(root == null){
            return;
        }
        ls.add(root.val);
        preorderTraversal(root.left, ls);
        preorderTraversal(root.right, ls);
    }
}

Method 2 是Iteration，维护一个stack，先压root，stack不空条件下每次循环, 先pop(), 然后压right child, 再压left child. 如此保证了出栈的顺序是preorder.

Time Complexity: O(n). Space: O(logn).

AC Java:

//Method 2: Iteration
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ls = new ArrayList<Integer>();
        if(root == null){
            return ls;
        }
        Stack<TreeNode> stk = new Stack<TreeNode>();
        stk.push(root);
        while(!stk.isEmpty()){
            TreeNode tn = stk.pop();
            ls.add(tn.val);
            if(tn.right != null){
                stk.push(tn.right);
            }
            if(tn.left != null){
                stk.push(tn.left);
            }
        }
        return ls;
    } 

Method 3 : 上面的方法是对的，但是为了统一记忆Preorder, Inorder, Postorder, 改写成了下面的Method 3. 

进栈的顺序都不变，与Binary Tree Inorder Traversal不同在于加ls的地方不同.

Time Complexity: O(n). Space: O(logn).

AC Java:

//Method 3: Iteration
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ls = new ArrayList<Integer>();
        Stack<TreeNode> stk = new Stack<TreeNode>();
        while(root!=null || !stk.isEmpty()){
            if(root != null){
                ls.add(root.val);
                stk.push(root);
                root = root.left;
            }else{
                root = stk.pop();
                root = root.right;
            }
        }
        return ls;
    }

Method 4: Morris Traversal

也是借鉴了Morris Traversal, 与Binary Tree Inorder Traversal相似，唯一不同就是加ls的时机不同。

Time Complexity: O(n). Space: O(1).

AC Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        TreeNode cur = root;
        TreeNode pre = null;
        while(cur != null){
            if(cur.left == null){
                res.add(cur.val);
                cur = cur.right;
            }else{
                pre = cur.left;
                while(pre.right != null && pre.right != cur){
                    pre = pre.right;
                }
                
                if(pre.right == null){
                    res.add(cur.val);
                    pre.right = cur;
                    cur = cur.left;
                }else{
                    pre.right = null;
                    cur = cur.right;
                }
            }
        }
        return res;
    }
}

 跟上Verify Preorder Sequence in Binary Search Tree, Subtree of Another Tree.