二叉树的递归遍历

[题目](144. 二叉树的前序遍历)

[题目](145. 二叉树的后序遍历)

[题目](94. 二叉树的中序遍历)

重点

确定递归函数的参数和返回值，确定终止条件，确定单层递归的逻辑

代码实现（递归）

// 前序
class Solution {
public:
    void traversal(TreeNode* cur, vector<int>& vec) {
        if (cur == NULL)
            return;
        vec.push_back(cur->val);
        traversal(cur->left, vec);
        traversal(cur->right, vec);
    }
    
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> result;
        traversal(root, result);
        return result;
    }
};
// 后序
class Solution {
public:
    void traversal(TreeNode* cur, vector<int>& vec) {
        if (cur == NULL)
            return;
        traversal(cur->left, vec);
        traversal(cur->right, vec);
        vec.push_back(cur->val);
    }
    
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> result;
        traversal(root, result);
        return result;
    }
};
// 中序
class Solution {
public:
    void traversal(TreeNode* cur, vector<int>& vec) {
        if (cur == NULL)
            return;
        traversal(cur->left, vec);
        vec.push_back(cur->val);
        traversal(cur->right, vec);
    }
    
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        traversal(root, result);
        return result;
    }
};

代码实现（迭代）

// 前序
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> st;
        vector<int> result;
        if (root == NULL)
            return result;
        st.push(root);
        while (!st.empty()) {
            TreeNode* node = st.top();
            st.pop();
            result.push_back(node->val);
            if (node->right)
                st.push(node->right);
            if (node->left)
                st.push(node-left);
        }
        return result;
    }
};
// 后序
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        stack<TreeNode*> st;
        vector<int> result;
        if (root == NULL)
            return result;
        st.push(root);
        while (!st.empty()) {
            TreeNode* node = st.top();
            st.pop();
            result.push_back(node->val);
            if (node->left)
                st.push(node-left);
            if (node->right)
                st.push(node->right);
        }
        reverse(result.begin(), result.end());
        return result;
    }
};
// 中序
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> st;
        TreeNode* cur = root;
        while (cur != NULL || !st.empty()) {
            // 指针来访问节点，访问到最底层
            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;
    }
};