• [LeetCode] 107. Binary Tree Level Order Traversal II 二叉树层序遍历 II


    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],

        3
       / 
      9  20
        /  
       15   7

    return its bottom-up level order traversal as:

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

    102. Binary Tree Level Order Traversal 的变形,只是最后输出的顺序变了,这题是从最后一层开始。

    Python:

    class TreeNode(object):
        def __init__(self, x):
            self.val = x
            self.left = None
            self.right = None
    
    
    class Solution(object):
        def levelOrderBottom(self, root):
            """
            :type root: TreeNode
            :rtype: List[List[int]]
            """
            if root is None:
                return []
    
            result, current = [], [root]
            while current:
                next_level, vals = [], []
                for node in current:
                    vals.append(node.val)
                    if node.left:
                        next_level.append(node.left)
                    if node.right:
                        next_level.append(node.right)
                current = next_level
                result.append(vals)
    
            return result[::-1] 

    C++:Iteration

    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;
        }
    };
    

    C++:Recursion

    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);
        }
    };  

      

    类似题目:

    [LeetCode] 102. Binary Tree Level Order Traversal 二叉树层序遍历

    [LeetCode] 199. Binary Tree Right Side View 二叉树的右侧视图

    All LeetCode Questions List 题目汇总

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/8606840.html
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