(LeetCode 面试经典 150 题) 103. 二叉树的锯齿形层序遍历(广度优先搜索bfs)

题目：103. 二叉树的锯齿形层序遍历

思路：广度优先搜索bfs，时间复杂度0(n)。

C++版本：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        if(root==nullptr) return ans;
        queue<TreeNode *> qu;
        qu.push(root);
        // 当前层节点数不为0
        while(qu.size()>0){
        	// 当前层节点数
            int n=qu.size();
            vector<int> v;
            for(int i=0;i<n;i++){
                TreeNode * tmp=qu.front();
                qu.pop();
                v.push_back(tmp->val);
                if(tmp->left!=nullptr) qu.push(tmp->left);
                if(tmp->right!=nullptr) qu.push(tmp->right);
            }
            if(ans.size()%2==1) reverse(v.begin(),v.end());
            ans.push_back(v);
        }
        return ans;
    }
};

JAVA版本：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> ans=new ArrayList<>();
        if(root==null) return ans;
        Queue<TreeNode> qu=new LinkedList<>();
        qu.add(root);
        
        while(qu.size()>0){
            int n=qu.size();
            List<Integer> v=new ArrayList<>();
            for(int i=0;i<n;i++){
                TreeNode  tmp=qu.poll();
                v.add(tmp.val);
                if(tmp.left!=null) qu.add(tmp.left);
                if(tmp.right!=null) qu.add(tmp.right);
            }
            if(ans.size()%2==1) Collections.reverse(v);
            ans.add(v);
        }
        return ans;
    }
}

GO版本：

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func zigzagLevelOrder(root *TreeNode) [][]int {
    ans:=[][]int{}
    if root==nil {return ans}
    qu:=[]*TreeNode{root}

    for len(qu)>0 {
        n:=len(qu)
        v:=make([]int,n)
        for i:=0;i<n;i++ {
            tmp:=qu[0]
            qu=qu[1:]
            if len(ans)%2==0 {
                v[i]=tmp.Val
            }else{
                v[n-i-1]=tmp.Val
            }
            if tmp.Left!=nil {
                qu=append(qu,tmp.Left)
            }
            if tmp.Right!=nil {
                qu=append(qu,tmp.Right)
            }
        }
        ans=append(ans,v)
    }
    return ans

}