Leetcode Binary Tree

最新推荐文章于 2025-12-06 07:43:37 发布

原创最新推荐文章于 2025-12-06 07:43:37 发布 · 77 阅读

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#leetcode #算法 #数据结构

本文介绍了如何使用递归和迭代方法来遍历二叉树，包括前序、中序和后序遍历。在递归方法中，通过反转子树顺序实现了前序和后序遍历。而在迭代方法中，利用栈进行操作，特别是后序遍历需要在遍历完成后反转结果数组。

二叉树的递归遍历

preorder(Leetcode 144)

public:
    vector<int> res;
    void reverse(TreeNode *cur){
        if(cur == NULL) return;
        res.push_back(cur->val);
        reverse(cur->left);
        reverse(cur->right);
    }
    vector<int> preorderTraversal(TreeNode* root) {
        reverse(root);
        return res;   
    }

postorder(Leetcode 145)

class Solution {
public:
    vector<int> res;
    void reverse(TreeNode *cur){
        if(cur == NULL) return;
        reverse(cur->left);
        reverse(cur->right);
        res.push_back(cur->val);
    }
    vector<int> postorderTraversal(TreeNode* root) {
        reverse(root);
        return res;
    }
};

3. Inorder(Leetcode 94)

class Solution {
public:
    vector<int> res;
    void reverse(TreeNode *cur){
        if(cur == NULL) return;

        reverse(cur->left);
        res.push_back(cur->val);
        reverse(cur->right);
    }
    vector<int> inorderTraversal(TreeNode* root) {
        reverse(root);
        return res;
    }
};

2. 迭代

preorder

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

postorder

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stk;

        if(root == NULL) return res;
        stk.push(root);

        while(!stk.empty()){
            TreeNode *cur = stk.top();
            res.push_back(cur->val);
            stk.pop();
            if(cur->left) stk.push(cur->left);
            if(cur->right) stk.push(cur->right);
            
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

inorder

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

        return res;
    }
};