[LeetCode] Binary Tree Level Order Traversal II 二叉树层序遍历之二

更新时间：2022-09-13 22:45:13

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

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

return its bottom-up level order traversal as:

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

从底部层序遍历其实还是从顶部开始遍历，只不过最后存储的方式有所改变，可以参见我之前的博文 http://www.cnblogs.com/grandyang/p/4051321.html。代码如下：

解法一：

// Iterative

class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > res;
        if (root == NULL) return res;

        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            vector<int> oneLevel;
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                TreeNode *node = q.front();
                q.pop();
                oneLevel.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            res.insert(res.begin(), oneLevel);
        }
        return res;
    }
};

下面我们来看递归的解法，核心就在于我们需要一个二维数组，和一个变量level，当level递归到上一层的个数，我们新建一个空层，继续往里面加数字，参见代码如下：

解法二：

// Recurive

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int> > res;
        levelorder(root, 0, res);
        return vector<vector<int> > (res.rbegin(), res.rend());
    }
    void levelorder(TreeNode *root, int level, vector<vector<int> > &res) {
        if (!root) return;
        if (res.size() == level) res.push_back({});
        res[level].push_back(root->val);
        if (root->left) levelorder(root->left, level + 1, res);
        if (root->right) levelorder(root->right, level + 1, res);
    }
};

本文转自博客园Grandyang的博客，原文链接：[LeetCode] Binary Tree Level Order Traversal II 二叉树层序遍历之二

，如需转载请自行联系原博主。

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[LeetCode] Binary Tree Level Order Traversal II 二叉树层序遍历之二

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