[LintCode] 66 Binary Tree Preorder Traversal 解题报告

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



Example
Given:

    1
   / \
  2   3
 / \
4   5
return [1,2,4,5,3].



Challenge
Can you do it without recursion?



思路
解法一：
Traverse


Code

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        
        ArrayList<Integer> result = new ArrayList<>();
        traverse(root, result);
        
        return result;
    }
    
    public void traverse(TreeNode root, ArrayList<Integer> result) {
        if (root == null) {
            return;
        }
        
        result.add(root.val);
        traverse(root.left, result);
        traverse(root.right, result);
    }
}

解法二：

Divide & Conquer

Code

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        
        ArrayList<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        
        
        ArrayList<Integer> left = preorderTraversal(root.left);
        ArrayList<Integer> right = preorderTraversal(root.right);
        
        result.add(root.val);
        result.addAll(left);
        result.addAll(right);
        
        return result;
    }
}

解法三：

Iterative

Code

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        
        ArrayList<Integer> result = new ArrayList<>();
        
        if (root == null) {
            return result;
        }
        
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            result.add(node.val);
            
            if (node.right != null) {
                stack.push(node.right);
            }
            if (node.left != null) {
                stack.push(node.left);
            }
        }
        
        return result;
    }
}