数据结构与算法（考研代码部分）

Posted 2020-12-02 jiangxiaobin1996

二叉树的遍历：

前序遍历（递归）： LeetCode 144

/**
 * 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<int> vec;
    vector<int> preorderTraversal(TreeNode* root) {
        if(root!=NULL){
            vec.push_back(root->val);
            preorderTraversal(root->left);
            preorderTraversal(root->right);
        }
        return vec;
    }
};

非递归：

/**
 * 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<int> preorderTraversal(TreeNode* root) {
        vector<int> vec;
        stack<TreeNode*> s;
        while(root || !s.empty()){
            if(root!=NULL){
                s.push(root);
                vec.push_back(root->val);
                root=root->left;
            }else{
                root=s.top();
                s.pop();
                root=root->right;
            }
        }
        return vec;
    }
};

中序遍历（递归）：LeetCode 94

/**
 * 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<int> vec;
    vector<int> inorderTraversal(TreeNode* root) {
        if(root){
            inorderTraversal(root->left);
            vec.push_back(root->val);
            inorderTraversal(root->right);
        }
        return vec;
    }
};

非递归：

/**
 * 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<int> inorderTraversal(TreeNode* root) {
       vector<int> vec;
        stack<TreeNode*> s;
        while(root || !s.empty()){
            if(root){
                s.push(root);
                root=root->left;
            }else{
                root=s.top();
                s.pop();
                vec.push_back(root->val);
                root=root->right;
            }
        }
        return vec;
    }
};

　　后序遍历（递归）： LeetCode 145

/**
 * 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<int> vec;
    vector<int> postorderTraversal(TreeNode* root) {
        if(root){
            postorderTraversal(root->left);
            postorderTraversal(root->right);
            vec.push_back(root->val);
        }
        return vec;
    }
};

　　非递归：（大致解析：如果该结点的左右子树都为空或者左右子树中的一个为上一次访问的结点即该结点的右结点已经访问过了，则输出。否则将该结点的右结点和左结点压入栈中）

PS：这是其中一种解法，还有一种解法是对每一个结点增加一个isFirst的属性来判断该结点是否是第一次访问，如果是第二次访问则输出。否则访问该结点的右结点。

/**
 * 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<int> postorderTraversal(TreeNode* root) {
        vector<int> vec;
        stack<TreeNode*> s;
        TreeNode* pre=NULL;
        if(root==NULL) return vec;
        else s.push(root);
        while(!s.empty()){
            root=s.top();
            if((root->left==NULL&&root->right==NULL) || (pre!=NULL&&(pre==root->left||pre==root->right))){
                s.pop();
                vec.push_back(root->val);
                pre=root;
            }else{
                if(root->right) s.push(root->right);
                if(root->left) s.push(root->left);
            }
        }
        return vec;
    }
};