Leetcode 102. Binary Tree Level Order Traversal

最新推荐文章于 2025-09-11 19:19:55 发布

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最新推荐文章于 2025-09-11 19:19:55 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，并提供了两种不同的实现方案：一种使用递归的方式，另一种则利用栈来实现。这两种方法都能有效地按层次顺序返回二叉树节点的值。

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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

/**
 * 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> > ret;
        reReadTree(root, 0, ret);
        return ret; 
    }

private:
    void reReadTree(TreeNode* root, size_t id, vector<vector<int> >& ret) {
        if (root == nullptr) return;
        if (id < ret.size()) 
            ret[id].emplace_back(root->val);
        else 
            ret.emplace_back(vector<int>{root->val});
        reReadTree(root->left, id + 1, ret);
        reReadTree(root->right, id + 1, ret);

    }
};

/**
 * 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) {
        stack<tuple<TreeNode*, size_t> >s;

        vector<vector<int> > ret;

        TreeNode* p = root;
        int id = 0;
        while (p != nullptr || !s.empty()) {
            if (p != nullptr) {
                if (id < ret.size())
                    ret[id].emplace_back(p->val);
                else 
                    ret.emplace_back(vector<int>{ p->val });
                s.push(make_tuple(p, id));
                p = p->left;
                ++id;
            } else {
                tuple<TreeNode*, size_t> t = s.top();
                p = get<0>(t);
                id = get<1>(t);
                s.pop();
                p = p->right;
                ++id;
            }
        }

        return ret;
    }
};