(C++)LeetCode#103. Binary Tree Zigzag Level Order Traversal

• Title: Zigzag print binary tree
• Difficulty: Hard
• Ideas: 1. A variable height is needed to indicate the printing direction (left to right or right to left)
  2. A variable numOfChild is used to indicate how many nodes there are in this layer
• Code:
  Implemented using a deque

/**
 * 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>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> ivec;
        if (root == NULL) {
            return ivec;
        }
        deque<TreeNode *> deq;
        deq.push_back(root);
        int height = 0;
        int numOfChild = 1;

        while(!deq.empty()) {
            vector<int> vec;
            for(int i = 0; i < numOfChild; i++) {
                TreeNode *node;
                //从左到右      
                if(height%2 == 0){
                    node = deq.back();
                    deq.pop_back();
                    if(node->left != NULL){
                        deq.push_front(node->left);
                    }
                    if(node->right != NULL){
                        deq.push_front(node->right);
                    }

                }else{
                    node  = deq.front();
                    deq.pop_front();
                    if(node->right != NULL){
                        deq.push_back(node->right);
                    }
                    if(node->left != NULL){
                        deq.push_back(node->left);
                    }
                }
                vec.push_back(node->val);
            }
            height++;
            numOfChild = deq.size();
            ivec.push_back(vec);
        }
        return ivec;
    }
};