力扣--二叉树的遍历题

这几题都是关于二叉树的遍历，前序中序后序，虽然比较基础，但还是做个笔记吧。代码都是官解的代码。

【前序遍历】

144. 二叉树的前序遍历
前序遍历官解

【递归法代码】

/**
 * 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:
    void preorder(TreeNode *root, vector<int> &res) {
        if (root == nullptr) {
            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;
    }
};

【迭代法代码】

class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if (root == nullptr) {
            return res;
        }

        stack<TreeNode*> stk;
        TreeNode* node = root;
        while (!stk.empty() || node != nullptr) {
            while (node != nullptr) {
                res.emplace_back(node->val);
                stk.emplace(node);
                node = node->left;
            }
            node = stk.top();
            stk.pop();
            node = node->right;
        }
        return res;
    }
};

【后序遍历】

145. 二叉树的后序遍历
后序遍历官解

【递归法代码】

class Solution {
public:
    void postorder(TreeNode *root, vector<int> &res) {
        if (root == nullptr) {
            return;
        }
        postorder(root->left, res);
        postorder(root->right, res);
        res.push_back(root->val);
    }

    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> res;
        postorder(root, res);
        return res;
    }
};

【迭代法代码】

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> res;
        if (root == nullptr) {
            return res;
        }

        stack<TreeNode *> stk;
        TreeNode *prev = nullptr;
        while (root != nullptr || !stk.empty()) {
            while (root != nullptr) {
                stk.emplace(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            if (root->right == nullptr || root->right == prev) {
                res.emplace_back(root->val);
                prev = root;
                root = nullptr;
            } else {
                stk.emplace(root);
                root = root->right;
            }
        }
        return res;
    }
};

【中序遍历】

【递归法】

class Solution {
public:

    void inorder(TreeNode*root, vector<int>& res){
        if(root==nullptr){
            return ;
        }
        inorder(root->left,res);
        res.push_back(root->val);
        inorder(root->right,res);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        inorder(root,res);
        return res;
    }
};

【迭代法】

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stk;
        while (root != nullptr || !stk.empty()) {
            while (root != nullptr) {
                stk.push(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            res.push_back(root->val);
            root = root->right;
        }
        return res;
    }
};