missjing/leetcode

Switch branches/tags
Nothing to show
Fetching contributors…
Cannot retrieve contributors at this time
74 lines (67 sloc) 1.5 KB
 problem: 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,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its zigzag level order traversal as: [ [3], [20,9], [15,7] ] confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ. ---------------------------------------------------------------------------------------- solution: ////////////////// struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; //两个栈(curLevel, nextLevel)来模拟，并用一个变量记录是l-to-r或r-to-l vector > zigzagLevelOrder(TreeNode *root) { vector > ret; if(root == NULL) return ret; stack curLevel, nextLevel; bool isLToR = true; curLevel.push(root); vector tmp; while(!curLevel.empty()) { TreeNode *curNode = curLevel.top(); curLevel.pop(); if(curNode) { tmp.push_back(curNode->val); if(isLToR) { if(curNode->left) nextLevel.push(curNode->left); if(curNode->right) nextLevel.push(curNode->right); } else { if(curNode->right) nextLevel.push(curNode->right); if(curNode->left) nextLevel.push(curNode->left); } } if(curLevel.empty()) { ret.push_back(tmp); tmp.clear(); isLToR = !isLToR; swap(curLevel, nextLevel); } } return ret; }