Binary Tree Level Order Traversal

初阶：

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by
level).
For example: Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

进阶。

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right,
level by level from leaf to root).
For example: Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:

[
[15,7]
[9,20],
[3],
]

进阶2。

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]
]

思路：

有两种方式实现，一个是采用队列来模拟实现。另外一种是采用递归。

先看一种：

//如果不考虑层次，只用一个vec表示，比较简单。vector<char> leverOrder2(TreeNode *root){vector<char> result;if(root == NULL)return result;queue<TreeNode*> q;q.push(root);while(!q.empty()){TreeNode *node = q.front();q.pop();result.push_back(node->data);if(node->left != NULL)q.push(node->left);if(node->right != NULL)q.push(node->right);}return result;}//如果考虑层次，需要用vec<vec<char>>来记录。vector<vector<char>> leverOrder3(TreeNode *root){vector<vector<char>> result;if(root == NULL)return result;queue<TreeNode *> q;q.push(root);vector<char> tmp;int count = 1;while(!q.empty()){tmp.clear();int levelCount = 0;for(int i=0;i<count;i++)//遍历当前层的所有节点{TreeNode *node = q.front();q.pop();tmp.push_back(node->data);//同一层的左右子节点也在同一层上，将新一层的节点插入到队列中if(node->left != NULL){q.push(node->left);++levelCount; //计算新一层的节点数目}if(node->right != NULL){q.push(node->right);++levelCount; //计算新一层的节点数目}}count = levelCount;//得出新的一层的节点数目result.push_back(tmp);}return result;}

在看用递归来实现：

#include <iostream>#include <vector>using namespace std;struct TreeNode{char data;TreeNode *left;TreeNode *right;TreeNode(char x) : data(x), left(NULL), right(NULL) {}};//思路。采用递归的方式。用一个二维数组表示,一维表示层，另外一维表示当前层的值。void levelOrder(vector<vector<char>> &ret,int dep,TreeNode *root){if(root == NULL){return;}if(ret.size()>dep)//如果层数大于树的深度时{ret[dep].push_back(root->data);}else{vector<char> a;a.push_back(root->data);ret.push_back(a);}levelOrder(ret,dep+1,root->left);levelOrder(ret,dep+1,root->right);}//采用递归的方式，和上面的区别在于，加一个flag记录是从左到右还是从右到左，每一层结束翻转一下void zigzagLevelOrder(vector<vector<char>> &ret,int dep,TreeNode *root,bool left_to_right){if(!root) return;if(ret.size()>dep){if(left_to_right){ret[dep].push_back(root->data);}else{ret[dep].insert(ret[dep].begin(),root->data);}}else{vector<char> a;a.push_back(root->data);ret.push_back(a);}zigzagLevelOrder(ret,dep+1,root->left,!left_to_right);zigzagLevelOrder(ret,dep+1,root->right,!left_to_right);}//创建一个二叉树void buildTree2(TreeNode **T){char ch;cin>>ch;if(ch == '#'){*T = NULL;}else{*T = new TreeNode(ch);printf("请输入%c的左孩子：",ch);buildTree2(&((*T)->left));printf("请输入%c的右孩子：",ch);buildTree2(&((*T)->right));}}//先序遍历void preTraverse(TreeNode *root){if(root == NULL) return;printf("%c ",root->data);if(root->left) preTraverse(root->left);if(root->right) preTraverse(root->right);}void main(){TreeNode *T;buildTree2(&T);vector<vector<char>> ret;//levelOrder(ret,0,T);zigzagLevelOrder(ret,0,T,true);for(int i=0;i<ret.size();i++){vector<char> a = ret[i];for(int j=0;j<a.size();j++){printf("%c ",a[j]);}printf("\n");}//preTraverse(T);//printf("\n");}