push(root); 28 tree_queue.push(NULL); 29 int nLevelCount = 1; 30 while (true) { 31 TreeNode *pTemp = tree_queue.front(); 32 tree_queue.pop(); 33 if (pTemp == NULL) { 34 if (nLevelCount%2 == 0) { //if the num of
Given a binary tree, return thezigzag level ordertraversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree[3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its zigzag level order t...
Given a binary tree, return thezigzag level ordertraversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree{3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its zigzag level order travers...
103. Binary Tree Zigzag Level Order Traversal Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree [3,9,20,null,null,15,7], 代码语言:...
offer(level); treeNode.offer(curNode.right); nodeLevel.offer(level); } } return ans; } 第二种方案 把102 题 的解释贴过来。 我们在 while 循环中加一个 for 循环,循环次数是循环前的队列中的元素个数即可,使得每次的 while 循环出队的元素都是同一层的元素。 for 循环结束也就意味着当前层结束了,...
Binary Tree Zigzag Level Order Traversal Given a binary tree, return thezigzag level ordertraversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree{3,9,20,#,#,15,7},...
103.binary-tree-zigzag-level-order-traversal Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree [3,9,20,null,null,15,7],...
levelTraversal(root->right, level+1, !leftToRight, result); } }; 2. 迭代解法 // 迭代解法 classSolution{ public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { vector<vector<int> > result; if(!root)returnresult;
103. Binary Tree Zigzag Level Order Traversal 58b6ace olsen-blue reviewed Mar 2, 2025 View reviewed changes 103/step3.cpp vector<TreeNode*> current_level_nodes = {root}; while (!current_level_nodes.empty()) { vector<TreeNode*> next_level_nodes; level_ordered_vals.push_back({...
103. Binary Tree Zigzag Level Order Traversal.md node = nodes[i] values.append(node.val) first, second = access_orders[level % 2] first_node = getattr(node, first) nodchip Feb 27, 2024 リフレクション的な機能は極力使用しないほうが良いと思います。理由は、コンパイル時の静...