leetcode-102-Binary Tree Level Order Traversal

最新推荐文章于 2024-06-25 11:36:41 发布

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最新推荐文章于 2024-06-25 11:36:41 发布

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分类专栏：算法刷题文章标签： leetcode 二叉树层级遍历

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本文介绍了一种实现二叉树层次遍历的方法，并通过具体的C++代码示例展示了如何构造二叉树并进行层次遍历，最后输出各层节点的值。

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#include <iostream>
#include <vector>
using namespace std;
//Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> myVec;
        vector<TreeNode *> levVec;
        if (!root) {
            return myVec;
        }
        levVec.push_back(root);
        myVec.push_back(vector<int>(1, root->val));
        while (true) {
            vector<TreeNode *> nextVec;
            vector<int> numVec;
            for (int i = 0; i < levVec.size(); i++) {
                if (levVec[i]->left) {
                    nextVec.push_back(levVec[i]->left);
                    numVec.push_back(levVec[i]->left->val);
                }
                if (levVec[i]->right) {
                    nextVec.push_back(levVec[i]->right);
                    numVec.push_back(levVec[i]->right->val);
                }
            }
            if (nextVec.size() == 0) {
                break;
            }
            levVec = nextVec;
            myVec.insert(myVec.begin(), numVec);
        }
        return myVec;
    }
};

int main(int argc, const char * argv[]) {
    TreeNode three = TreeNode(3);
    TreeNode nine = TreeNode(9);
    TreeNode twenty = TreeNode(20);
    TreeNode fifteen = TreeNode(15);
    TreeNode seven = TreeNode(7);
    three.left = &nine;
    three.right = &twenty;
    twenty.left = &fifteen;
    twenty.right = &seven;
    Solution s;
    vector<vector<int>> res = s.levelOrder(&three);
    for (int i = 0; i < res.size(); i++) {
        for (int j = 0; j < res[i].size(); j++) {
            cout << res[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}