二叉树的前中后序遍历

144. 二叉树的前序遍历

方法：递归

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

$时间复杂度O(n)，空间复杂度O(n)

方法：迭代

/**
 * Definition for a binary tree node.
 * 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) {}
 * };
 */
class Solution {
public:
    vector preorderTraversal(TreeNode* root) {
        stack st;
        vector res;
        if (root != NULL) st.push(root);

        while (!st.empty()) {
            TreeNode* nod = st.top();
            if (nod != NULL) {
                st.pop();
                if (nod->right) st.push(nod->right);
                if (nod->left) st.push(nod->left);
                st.push(nod);
                st.push(NULL);
            } else {
                st.pop();
                nod = st.top();
                st.pop();
                res.push_back(nod->val);
            }
        }
        return res;
    }
};

$时间复杂度O(n)，空间复杂度O(n)

145. 二叉树的后序遍历

方法：递归

/**
 * Definition for a binary tree node.
 * 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) {}
 * };
 */
class Solution {
    vectorret;
    
    void dfs(TreeNode* root){
        if(root){
            dfs(root->left);
            dfs(root->right);
            ret.push_back(root->val);
        }
    }
public:
    vector postorderTraversal(TreeNode* root) {
        ret.clear();
        dfs(root);
        return ret;
    }
};

$时间复杂度O(n)，空间复杂度O(n)

方法：迭代

/**
 * Definition for a binary tree node.
 * 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) {}
 * };
 */
class Solution {
public:
    vector postorderTraversal(TreeNode* root) {
        stack st;
        vector res;
        if (root != NULL) st.push(root);

        while (!st.empty()) {
            TreeNode* nod = st.top();
            if (nod != NULL) {
                st.pop();
                st.push(nod);
                st.push(NULL);
                if (nod->right) st.push(nod->right);
                if (nod->left) st.push(nod->left);
            } else {
                st.pop();
                nod = st.top();
                st.pop();
                res.push_back(nod->val);
            }
        }
        return res;
    }
};

$时间复杂度O(n)，空间复杂度O(n)

class Solution {
private:
    void solve(TreeNode* cur, vector& vec) {
        if (cur == nullptr) return ;
        
        solve(cur->left, vec);
        vec.push_back(cur->val);
        solve(cur->right, vec);
    }
public:
    vector inorderTraversal(TreeNode* root) {
        vector vec;
        solve(root, vec);
        return vec;
    }
};

$时间复杂度O(n)，空间复杂度O(n)

方法：迭代

/**
 * Definition for a binary tree node.
 * 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) {}
 * };
 */
class Solution {
public:
    vector inorderTraversal(TreeNode* root) {
        stack st;
        vector res;
        if (root != NULL) st.push(root);

        while (!st.empty()) {
            TreeNode* nod = st.top();
            if (nod != NULL) {
                st.pop();
                if (nod->right) st.push(nod->right);
                st.push(nod);
                st.push(NULL);
                if (nod->left) st.push(nod->left);
            } else {
                st.pop();
                nod = st.top();
                st.pop();
                res.push_back(nod->val);
            }
        }
        return res;
    }
};

$时间复杂度O(n)，空间复杂度O(n)