[LeetCode] binary-tree-preorder-traversal

最新推荐文章于 2019-05-12 14:29:37 发布

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本文介绍了一种实现二叉树前序遍历的非递归算法，并提供了一个具体的Java实现示例。该算法利用栈来存储节点，确保了遍历顺序符合前序遍历的要求。

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

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

return[1,2,3].

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

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

public class Tree {
	public class TreeNode{
		int val;
		TreeNode left;
		TreeNode right;
		TreeNode(int x){val = x;}
	}
	
	public TreeNode init(){
		TreeNode root = new TreeNode(1);
		//1初始化根
		root.left = null;
		root.right = new TreeNode(2);
		
		//2初始化根的右子树
		root.right.left = new TreeNode(3);
		root.right.right = null;
		
		//返回根节点
		return root;
	}
	
	public static void main(String[] args){
		TreeNode root = new Tree().init();
		System.out.println(preorderTravelsal(root));
		
	}
	
	public static ArrayList<Integer> preorderTravelsal(TreeNode root){
		ArrayList<Integer> list = new ArrayList<>();
		if (root == null) {
			return list;
		}
		//使用栈结构进行存储，因为是先进后出，所以除了一开始根入栈外，后续的入栈顺序是右→左，这样弹出时即为前序遍历的顺序
		Stack<TreeNode> stack = new Stack<>();
		stack.push(root);
		while(!stack.isEmpty()){
			TreeNode t = stack.pop();
			list.add(t.val);
			if (t.right != null) {
				stack.push(t.right);
			}
			if (t.left != null) {
				stack.push(t.left);
			}
		}
		return list;
	}
}