Binary Tree Preorder Traversal

https://oj.leetcode.com/problems/binary-tree-preorder-traversal/

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?

解题思路：

很简单，递归。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List list = new ArrayList<Integer>();
        if(root == null){
            return list;
        }
        list.add(root.val);
        list.addAll(preorderTraversal(root.left));
        list.addAll(preorderTraversal(root.right));
        return list;
    }
}

但原题提出不用递归的方法，即遍历。使用stack，遇到一个节点，首先处理该节点，即list.add(root.val)。如果右节点不为空，将右节点推入stack，留在后面弹出使用。这时如果左节点不为空，下移到左节点，处理左子树。如果左节点为空，从stack中弹出最上面的节点作为下一个节点。

循环上述过程，直到当前处理节点为空。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List list = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while(root!= null){
            list.add(root.val);
            if(root.right != null){
                stack.push(root.right);
            }
            if(root.left != null){
                root = root.left;
            }else{
                if(!stack.empty()){
                    root = stack.pop();
                }else{
                    root = null;
                }
            }
        }
        return list;
    }
}

 

下面是另一个比较简单的版本，也是推荐的。

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