102 - Binary Tree Level Order Traversal(DFS)

最新推荐文章于 2021-11-24 23:56:30 发布

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本文介绍了一种解决二叉树层次遍历问题的方法，通过深度优先遍历算法实现节点值从左到右、逐层返回。以示例形式展示了如何构造测试用例并调用算法。

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Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

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思路分析：

1、深度优先遍历算法

#include "stdafx.h"
#include <iostream>
#include <vector>
#include "BinNode.h"

using namespace std;

struct TreeNode
{
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


class Solution_102_BinaryTreeLevelOrderTraversal
{
private:
	vector<vector<int> > ret;
public:
	void solve(int dep, TreeNode *root)
	{
		if (root == NULL)
		{
			return;
		}


		if (ret.size() > dep)
		{
			ret[dep].push_back(root->val);
		}
		else
		{
			vector<int> a;
			a.push_back(root->val);
			ret.push_back(a);
		}

		solve(dep + 1, root->left);
		solve(dep + 1, root->right);
	}
	vector<vector<int>> levelOrder(TreeNode* root)
	{
		ret.clear();
		solve(0, root);

		return ret;
	}
};

测试用例的构造：

TreeNode *root = new TreeNode(3);

	TreeNode *leftson = new TreeNode(9);
	TreeNode *rightson = new TreeNode(20);
	root->left = leftson;
	root->right = rightson;

	TreeNode *leftson2 = new TreeNode(15);
	TreeNode *rightson2 = new TreeNode(7);
	rightson->left = leftson2;
	rightson->right = rightson2;
	

	Solution_102_BinaryTreeLevelOrderTraversal *s102 = new Solution_102_BinaryTreeLevelOrderTraversal;
	s102->levelOrder(root);