對二叉樹DFS\BFS的總結

DFS有遞歸與非遞歸兩種常見形式，BFS則通常為非遞歸的

本文使用

TreeNode.h

如下

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) {}
};
           

1. DFS

遞歸法

以前序遞歸法為例

c++

class Solution {
public:
	void preorder(TreeNode* root, vector<int> res){
		if (!root)  return;
		res.push_back(root->val);
		preorder(root->left, res); 
		preorder(root->right, res);
	}
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        preorder(root, res);
        return res;
    }
};
           

中序後序遞歸隻需改變

preorder

函數的語句順序即可

非遞歸形式

仍以前序DFS為例

c++

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> st;
        vector<int> result;
        st.push(root);
        while (!st.empty()) {
            TreeNode* node = st.top();                      // 中
            st.pop();
            if (node != NULL) result.push_back(node->val);
            else continue;
            st.push(node->right);                           // 右
            st.push(node->left);                            // 左
        }
        return result;
    }
};
           

2. BFS

c++

#include<TreeNode.h>
#include<vector>
#include<queue>
using namespace std;

class Solution {
public:
	vector<int> bfs(TreeNode* root) {
		vector<int> res;
		queue<TreeNode*> que;
		que.push(root);
		while (!que.empty()) {
			TreeNode* tmpNode = que.front();
			que.pop();
			if (tmpNode->left != nullptr) {
				que.push(tmpNode->left);
				res.push_back(tmpNode->left->val);
			}
			if (tmpNode->right != nullptr) {
				que.push(tmpNode->right);
				res.push_back(tmpNode->right->val);
			}
		}
		return res;
	}
};
           

1. 循環前初始化根結點進隊列
2. 循環開始出隊一個元素
3. 其後使左右孩子都進入隊列
4. 隊列不為空時循環往複

層序周遊

/**
 * 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>> levelOrder(TreeNode* root) {
        queue<TreeNode*> que;
        if (root != NULL) que.push(root);
        vector<vector<int>> res;
        while (!que.empty()) {
            int size = que.size();
            vector<int> vec;  //準備工作
            for(int i = 0; i < size; i++){  //内層才是明顯的bfs
                TreeNode* node = que.front();
                que.pop();
                vec.push_back(node->val);
                if (node->left) que.push(node->left);
                if (node->right) que.push(node->right);
            }
            res.push_back(vec);
        }
        return res;
    }
};