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,#,#,15,7}
,
3 / 9 20 / 15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
层次遍历,每一层加入一个vector<int> cur
当遍历到新一层时,将cur加入result,并清空,准备记录新一层数据。
以root为第0层,则奇数层的cur在加入result之前要reverse,变为从右往左。
以上描述涉及两个问题:
1、怎么知道当前在第几层?
2、怎么知道当前进入了新层?
解决:
1、为了判断遍历到的层数,设置一个Node结构,里面存放level。level从0起计数。
第i层的节点将子女进队列时,层数设为i+1。
2、设置一个变量lastLevel,记录上一个pop出来节点的层数。
如果当前pop出来节点的层数为lastLevel+1,即为到了新的层次。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ struct Node { TreeNode* tNode; int level; Node(TreeNode* newtNode, int newlevel): tNode(newtNode), level(newlevel) {} }; class Solution { public: vector<vector<int> > zigzagLevelOrder(TreeNode *root) { vector<vector<int> > ret; if(!root) return ret; // push root Node* rootNode = new Node(root, 0); queue<Node*> Nqueue; Nqueue.push(rootNode); vector<int> cur; int curlevel = 0; while(!Nqueue.empty()) { Node* frontNode = Nqueue.front(); Nqueue.pop(); if(frontNode->level > curlevel) { if(curlevel%2 == 1) reverse(cur.begin(), cur.end()); ret.push_back(cur); cur.clear(); curlevel = frontNode->level; } cur.push_back(frontNode->tNode->val); if(frontNode->tNode->left) { Node* leftNode = new Node(frontNode->tNode->left, frontNode->level+1); Nqueue.push(leftNode); } if(frontNode->tNode->right) { Node* rightNode = new Node(frontNode->tNode->right, frontNode->level+1); Nqueue.push(rightNode); } } if(curlevel%2 == 1) reverse(cur.begin(), cur.end()); ret.push_back(cur); return ret; } };