14代码随想录训练营day14|二叉树part01

1、二叉树的递归遍历

1.1、前序遍历

 144. 二叉树的前序遍历 - 力扣（LeetCode）

class Solution {
public:
    void ergodic(TreeNode* cur, vector& vec) {
        // 终止条件
        if (cur == NULL) return ;
        // 存根结点
        vec.push_back(cur->val);
        // 递归左子树
        ergodic(cur->left, vec);
        // 递归右子树
        ergodic(cur->right, vec);
    }

    vector preorderTraversal(TreeNode* root) {
        vector result;
        ergodic(root, result);
        return result;
    }
};

1.2、中序遍历

94. 二叉树的中序遍历 - 力扣（LeetCode）

class Solution {
public:
    void ergodic(TreeNode *cur, vector& vec) {
        if (cur == NULL) return ;
        ergodic(cur->left, vec);
        // 对比先序遍历的顺序
        vec.push_back(cur->val);
        ergodic(cur->right, vec);
    }
    vector inorderTraversal(TreeNode* root) {
        vector result;
        ergodic(root, result);
        return result;
    }
};

1.3、后序遍历

145. 二叉树的后序遍历 - 力扣（LeetCode）

class Solution {
public:
    void ergodic(TreeNode *cur, vector &vec) {
        if (cur == NULL) return ;
        ergodic(cur->left, vec);
        ergodic(cur->right, vec);
        vec.push_back(cur->val);
    }
    vector postorderTraversal(TreeNode* root) {
        vector result;
        ergodic(root, result);
        return result;
    }
};

2、迭代法

2.1、先序遍历

class Solution {
public:
    vector preorderTraversal(TreeNode* root) {
       // 用栈压入元素
       stack mystack;
       vector result;
       if (root == NULL) return result;
       mystack.push(root);
       while (!mystack.empty()) {
           TreeNode *temp = mystack.top();
           result.push_back(temp->val);
           mystack.pop();
           // 先压入right，后压入left，后进先出
           if (temp->right != NULL) mystack.push(temp->right);
           if (temp->left != NULL) mystack.push(temp->left);
       }
       return result;
    }
};

2.2、后序遍历

class Solution {
public:
    vector postorderTraversal(TreeNode* root) {
        stack mystack;
        vector result;
        if (root == NULL) return result;
        mystack.push(root);
        while (!mystack.empty()) {
            TreeNode* temp = mystack.top();
            mystack.pop();
            result.push_back(temp->val);
            if (temp->left != NULL) mystack.push(temp->left);
            if (temp->right != NULL) mystack.push(temp->right);
        }
        // reverse(result.begin(), result.end());
        for (int i = 0, j = result.size() - 1; i < j; i++, j--) {
            swap(result[i], result[j]);
        }
        return result;
    }
};

2.3、中序遍历

class Solution {
public:
    vector inorderTraversal(TreeNode* root) {
        vector result;
        stack st;
        // 设立指针
        TreeNode* cur = root;
        while (cur != NULL || !st.empty()) {
            if (cur != NULL) {
                st.push(cur);
                cur = cur->left;
            } else {
                cur = st.top();
                st.pop();
                result.push_back(cur->val);
                cur = cur->right;
            }
        }
        return result;
    }
};