107BinaryTreeLevelOrderTraversalII.org

[#C] 107 Binary Tree Level Order Traversal II - LeetCode

题目

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

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],[3]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

思路

层序遍历，和102题类似，先把每一层的元素加入到队列，计算每一层节点的个数，一次取出放到 List 第一元素的位置中。

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> levelList  = new ArrayList();
        // 注意 root 为空的判断，边界条件
        if (root == null) return levelList;
        Queue<TreeNode> q = new LinkedList();
        q.offer(root); 
     
        while (!q.isEmpty()) {
            int size = q.size();
            List<Integer> list = new ArrayList();
            for (int i = 0; i < size; i++) {
                TreeNode node = q.poll();
                list.add(node.val);
             
                if (node.left != null) q.offer(node.left);
                if (node.right != null) q.offer(node.right);
            }
            levelList.add(0, list);
        }
        return levelList;
    }
}

CPP

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        if (root == nullptr) 
            return res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            int size = q.size();
            vector<int> level;
            for (int i = 0; i < size; ++i) {
                auto node = q.front();
                q.pop();
                if (node->left != nullptr) q.push(node->left);
                if (node->right != nullptr) q.push(node->right);
                level.push_back(node->val);
            }
            res.insert(res.begin(), level);
        }
        return res;
    }
};