給定一個二叉樹，返回其節點值自底向上的層次遍歷。 （即按從葉子節點所在層到根節點所在的層，逐層從左向右遍歷）

例如：
給定二叉樹?[3,9,20,null,null,15,7],

3/ \9 20/ \15 7

返回其自底向上的層次遍歷為：

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

解法一：?

/*** Definition for a binary tree node.* 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) {if(!root) return {};queue<TreeNode *> qu;qu.push(root);vector<vector<int>> res;while(!qu.empty()){vector<int> temp;int n = qu.size();while(n--){TreeNode * s = qu.front();qu.pop();temp.push_back(s->val);if(s->left) qu.push(s->left);if (s->right) qu.push(s->right);}res.insert(res.begin(), temp);}return res; } };

解法二：

/*** Definition for a binary tree node.* 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>> result;vector<int> temp;queue<TreeNode*> TreeQueue;stack<vector<int>> TreeStack;unsigned long cursize = 0;if (root == NULL){return result;}TreeQueue.push(root);while (TreeQueue.size() != 0){cursize = TreeQueue.size();for (unsigned int index = 0; index < cursize; index++){TreeNode* indexnode = TreeQueue.front();TreeQueue.pop();temp.push_back(indexnode->val);if (indexnode->left != NULL){TreeQueue.push(indexnode->left);}if (indexnode->right != NULL){TreeQueue.push(indexnode->right);}}TreeStack.push(temp);temp.clear();}while (TreeStack.empty() != true){result.push_back(TreeStack.top());TreeStack.pop();}return result;} };

?

總結

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