LeetCode - Binary Tree Level Order Traversal II

最新推荐文章于 2026-01-05 22:36:07 发布

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#leetcode #binary tree #bfs

本文介绍了一种从叶子节点到根节点的层次遍历二叉树的方法，并提供了一个基于广度优先搜索（BFS）的解决方案。通过在每个层级结束时添加一个空指针来标记层级边界，实现从下到上的层次遍历。

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],
]

http://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

Solution:

Modified BFS. Add a NULL at the end of each level, so we can know that it reached to the end of this level.

https://github.com/starcroce/leetcode/blob/master/binary_tree_level_order_traversal_II.cpp

// 40 ms for 34 test cases
// same as version I except adding the reverse at the end
/**
 * 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;
        }
        queue<TreeNode *> q;
        q.push(root);
        q.push(NULL);
        vector<int> level;
        while(q.size() > 0) {
            TreeNode *tmp = q.front();
            q.pop();
            if(tmp != NULL) {
                level.push_back(tmp->val);
                if(tmp->left != NULL) {
                    q.push(tmp->left);
                }
                if(tmp->right != NULL) {
                    q.push(tmp->right);
                }
            } 
            else {
                res.push_back(level);
                level.erase(level.begin(), level.end());
                // previous level has been all poped, current level has been all pushed
                if(q.size() > 0) {
                    q.push(NULL);
                }
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};