[Leetcode] 107. Binary Tree Level Order Traversal II 解题报告

最新推荐文章于 2020-03-23 13:52:24 发布

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本文介绍了一种解决LeetCode 107题的方法，即从叶节点到根节点的层次遍历二叉树，并提供了详细的C++实现代码。

题目：

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,null,null,15,7],

return its bottom-up level order traversal as:

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

思路：

这道题目和Leetcode 102的唯一区别就是最后需要把结果翻转一下，具体请参加我的Leetcode 102解题报告。

代码：

/**
 * 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>> levelOrderBottom(TreeNode* root) {
        if (root == NULL) {
            return {};
        }
        vector<vector<int>> ret;
        queue<TreeNode*> q;
        q.push(root);
        q.push(NULL);       // separator
        vector<int> line;
        while (!q.empty()) {
            TreeNode* node = q.front();
            q.pop();
            if (node == NULL) {
                ret.push_back(line);
                line.clear();
                if (!q.empty()) {
                    q.push(NULL);
                }
            }
            else {
                line.push_back(node->val);
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
        }
        reverse(ret.begin(), ret.end());    // the only difference with Leetcode 102
        return ret;
    }
};