題目: LeetCode 144. 二叉樹的前序遍歷

給定一個二叉樹，返回它的 前序 遍歷。

示例:

輸入: [1,null,2,3] 1 2 / 3

輸出: [1,2,3] 進階: 遞迴演算法很簡單，你可以通過迭代演算法完成嗎？

遞迴演算法

import java.util.ArrayList;
import java.util.List;

/// 144. Binary Tree Preorder Traversal
/// https://leetcode.com/problems/binary-tree-preorder-traversal/description/
/// 二叉樹的前序遍歷
/// 時間複雜度: O(n), n為樹的節點個數
/// 空間複雜度: O(h), h為樹的高度
public class Solution144 {

    // Definition for a binary tree node.
    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        TreeNode(int x) { val = x; }
    }

    public List<Integer> preorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        preorderTraversal(root, res);
        return res;
    }

    private void preorderTraversal(TreeNode node, List<Integer> list){
        if(node != null){
            list.add(node.val);
            preorderTraversal(node.left, list);
            preorderTraversal(node.right, list);
        }
    }
}
 

非遞迴演算法

// Definition for a binary tree node.
public class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

// Classic Non-Recursive algorithm for preorder traversal
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution1 {

    public List<Integer> preorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while(!stack.empty()){
            TreeNode 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;
    }

}
 

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

// Another Classic Non-Recursive algorithm for preorder traversal
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution2 {

    public List<Integer> preorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        while(cur != null || !stack.isEmpty()){
            while(cur != null){
                res.add(cur.val);
                stack.push(cur);
                cur = cur.left;
            }

            cur = stack.pop();
            cur = cur.right;
        }
        return res;
    }
}
 

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

// Another Classic Non-Recursive algorithm for preorder traversal
// Time Complexity: O(n), n is the node number in the tree
// Space Complexity: O(h), h is the height of the tree
public class Solution3 {

    public List<Integer> preorderTraversal(TreeNode root) {

        ArrayList<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;

        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        while(cur != null || !stack.isEmpty()){
            if(cur != null){
                res.add(cur.val);
                stack.push(cur);
                cur = cur.left;
            }
            else{
                cur = stack.pop();
                cur = cur.right;
            }
        }
        return res;
    }
}