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  • 【Lintcode】070.Binary Tree Level Order Traversal II

    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).

    Example

    Given binary tree {3,9,20,#,#,15,7},

        3
       / 
      9  20
        /  
       15   7
    

     return its bottom-up level order traversal as:

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

    题解:

      迭代法,利用队列,不能用栈。

    Solution 1 ()

    class Solution {
    public:
        vector<vector<int>> levelOrderBottom(TreeNode *root) {
            vector<vector<int>> res;
            if (root == nullptr) {
                return res;
            }
            vector<int> v;
            queue<TreeNode*> q;
            q.push(root);
            q.push(nullptr);
            
            while (!q.empty()) {
                TreeNode* cur = q.front();
                q.pop();
                if(cur == nullptr) {
                    res.insert(res.begin(),v);
                    v.clear();
                    if(!q.empty()) {
                        q.push(nullptr);
                    }
                    continue;
                }
                v.push_back(cur->val);
                if(cur->left != nullptr) {
                    q.push(cur->left);
                }
                if(cur->right != nullptr) {
                    q.push(cur->right);
                }
            }
            
            return res;
        }
    };

    Solution 1.1 ()

    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> v;
                for (int i = 0; i < size; ++i) {
                    TreeNode* cur = q.front();
                    q.pop();
                    if(cur->left != nullptr) {
                        q.push(cur->left);
                    }
                    if(cur->right != nullptr) {
                        q.push(cur->right);
                    }
                    v.push_back(cur->val);
                }
                res.insert(res.begin(),v);
            }
            return res;
            
        }
    };

      递归

    Solution 2 ()

     

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  • 原文地址:https://www.cnblogs.com/Atanisi/p/6906905.html
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