[Leetcode 95] 103 Binary Tree Zigzag Level Order Traversal
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] ]
Analysis:
A little change to the original level-order reaversal code. Add a zz flag to indicate whether or not reverse the current level's elements.
Remember that after exit the while loop, an extra push_back operation is needed !
Code:
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > zigzagLevelOrder(TreeNode *root) { 13 // Start typing your C/C++ solution below 14 // DO NOT write int main() function 15 vector<vector<int> > res; 16 17 if (root == NULL) 18 return res; 19 20 vector<int> l; 21 22 queue<TreeNode*> q; 23 q.push(root); 24 25 TreeNode *senti = new TreeNode(-1); 26 senti->left = senti->right = NULL; 27 q.push(senti); 28 29 bool zz = false; 30 31 while (q.size() != 1) { 32 TreeNode *n = q.front(); 33 q.pop(); 34 35 if (n == senti) { // get the sentinal node 36 if (zz) reverse(l.begin(), l.end()); 37 38 res.push_back(l); 39 q.push(n); 40 zz = !zz; 41 l.clear(); 42 } else { 43 l.push_back(n->val); 44 45 if (n->left != NULL) q.push(n->left); 46 if (n->right != NULL) q.push(n->right); 47 } 48 } 49 if (zz) reverse(l.begin(), l.end()); 50 res.push_back(l); 51 52 return res; 53 } 54 };
