[leetcode 107]Binary Tree Level Order Traversal II

最新推荐文章于 2025-12-28 17:52:30 发布

原创最新推荐文章于 2025-12-28 17:52:30 发布 · 218 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #BFS #tree

[LeetCode 刷题] 专栏收录该内容

158 篇文章

订阅专栏

本文介绍了一种算法，用于实现二叉树的层次遍历并以从叶节点到根节点的方式返回各层级节点值。该算法使用了迭代方式，并记录每一层的节点值。

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]
]

/**
 * Definition for binary tree
 * 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) {
        vector<vector<int> > res;
        if (root == NULL) {
            return res;
        }
        vector<int> r;
        r.push_back(root->val);
        res.push_back(r);
        vector<TreeNode *> path;
        path.push_back(root);
        int count = 1;
        while (!path.empty()) {
            auto p = path.front();
            if (p->left) {
                path.push_back(p->left);
            }
            if (p->right) {
                path.push_back(p->right);
            }
            count --;
            path.erase(path.begin());
            if (count == 0) {
                vector<int> tmp;
                for (auto iter = path.begin(); iter != path.end(); iter ++) {
                    tmp.push_back((*iter)->val);
                }
                if (tmp.size()>0){
                     res.push_back(tmp);
                }
                count = path.size();
            }
        }
        reverse(res.begin(),res.end());
        return res;
        
    }
};