二叉树的层序遍历 BFS

二叉树的层序遍历，或者说是宽度优先便利，是经常考察的内容。

 

问题一：层序遍历二叉树并输出，直接输出结果即可，输出格式为一行。

 

#include <iostream>
#include <vector>
#include <deque>
#include <map>
#include <set>
#include <string>
#include <cstring>
#include <cstdlib>

using namespace std;

typedef struct BinTree{
	int data;
	struct BinTree *left;
	struct BinTree *right;
}BinTree;

/* 按照层序遍历方法遍历二叉树,使用一个队列来辅助 */
void BreadthFirst(BinTree *root){
	if(root == NULL)
		return;
	deque<BinTree *> q;
	q.push_back(root);

	BinTree *p;
	while(!q.empty()){
		p = q.front();
		q.pop_front();

		cout<<p->data;

		if(p->left)
			q.push_back(p->left);

		if(p->right)
			q.push_back(p->right);
	}
}

//测试
int main()
{
    /*  创建以下的树
            10
          /   \
        8      2
      /  \    /
    3     5  2
  */
  struct node *root = newNode(10);
  root->left        = newNode(8);
  root->right       = newNode(2);
  root->left->left  = newNode(3);
  root->left->right = newNode(5);
  root->right->left = newNode(2);
  iterativePreorder(root);
  return 0;
}

 

问题二： 按层输出二叉树，每一层单独输出为一行。

#include <vector>
#include <deque>
#include <queue>

using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    vector<int> rightSideView(TreeNode *root) {
        vector<int> ans;
        if(root == nullptr) return ans;
        queue<TreeNode* > que;
        que.push(root);

        TreeNode* curr;
        while(!que.empty()) {
            int cnt = que.size();
            for(int i = 0; i < cnt; i++) {
                curr = que.front();
                que.pop();
                if(curr->left) {
                    que.push(curr->left);
                }
                if(curr->right) {
                    que.push(curr->right);
                }
                cout << curr->val << " ";
            }
            ans.push_back(curr->val);
            cout << endl;
        }
        return ans;
    }
};

int main() {
        /*  创建以下的树
            10
          /   \
        8      2
      /  \    /
    3     5  2
        */

    Solution sln;

    TreeNode *root = new TreeNode(10);
    root->left        = new TreeNode(8);
    root->right       = new TreeNode(2);
    root->left->left  = new TreeNode(3);
    root->left->right = new TreeNode(5);
    root->right->left = new TreeNode(2);
    sln.rightSideView(root);

    return 0;
}

 LeetCode: Binary Tree Right Side View

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

You should return [1, 3, 4].

 

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> result;
        if(root == NULL) return result;

        deque<TreeNode *> q;
        q.push_back(root);

        TreeNode *current;
        while(!q.empty()) {
            int len = q.size();
            for(int i = 0; i < len; i++) {
                current = q.front();
                q.pop_front();
                if(current->left) {
                    q.push_back(current->left);
                }

                if(current->right) {
                    q.push_back(current->right);
                }
            }
            result.push_back(current->val);
        }
        return result;
    }
};

 同理：可以得到二叉树的左视图求解方式。

    vector<int> leftSideView(TreeNode *root)
    {
        vector<int> ans;
        if(root == NULL) return ans;
        queue<TreeNode* > que;
        que.push(root);

        TreeNode* curr;
        while(!que.empty()) {
            int cnt = que.size();

            for(int i = 0; i < cnt; i++) {
                curr = que.front();
                if( i == 0){
                    ans.push_back(curr->val);
                }
                que.pop();

                if(curr->left) {
                    que.push(curr->left);
                }
                if(curr->right) {
                    que.push(curr->right);
                }
            }
        }
        return ans;
    }