leetcode 94|144|145. Binary Tree Traversal 二叉树三种遍历

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本文详细介绍了二叉树的三种遍历方式：前序遍历、中序遍历及后序遍历。包括递归与非递归两种实现方法，并提供了C++代码示例。

94. Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

return [1,3,2].

中序遍历必背！
way-1:递归
way-2:非递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void inorder(vector<int> &result, TreeNode *p)  
    {
        if (p->left)    
            inorder(result, p->left);
            
        result.push_back(p->val);
        
        if (p->right)
            inorder(result, p->right); 
    }
    
    vector<int> inorderTraversal(TreeNode* root) 
    {
        //way-1
        /*
        vector<int> result;
        if (root)
            inorder(result, root);
        return result;
        */

        //way-2
        vector<int> ret;
        stack<TreeNode *> s;
        TreeNode * p = root;
        
        while (p != NULL || !s.empty())
        {
            while ( p != NULL)
            {
                s.push(p);
                p = p->left;
            }
            if (!s.empty())
            {
                p = s.top();
                ret.push_back(p->val);
                s.pop();
                p = p->right;
            }
        } 
        return ret;
    }
};

144. Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

return [1,2,3].

前序遍历

way-1 : 递归
way-2: 非递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void preorderTraversal1(vector<int> &ret,TreeNode* root)
    {
        if (!root)
            return;
        ret.push_back(root->val);
        preorderTraversal1(ret, root->left);
        preorderTraversal1(ret, root->right);
    }
    
    vector<int> preorderTraversal(TreeNode* root) 
    {
        
        //way-1 : 递归
        /*
        vector<int> ret;
        preorderTraversal1(ret, root);
        return ret;
        */
        
        //way-2: 非递归
        vector<int> ret;
        stack<TreeNode *> haspoint;
        if (root)
            haspoint.push(root);
      
        while(!haspoint.empty())
        {
            TreeNode* p = haspoint.top();
            haspoint.pop();
            ret.push_back(p->val);
            
            if (p->right)
                haspoint.push(p->right);
            if (p->left)
                haspoint.push(p->left);
        }
        return ret;
    }
};

145. Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [3,2,1].

后序遍历

1、递归

2、非递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void recursive(TreeNode* p,vector<int> & ret)
    {
        if(!p)   
            return;
        recursive(p->left,ret);
        recursive(p->right,ret);       
        ret.push_back(p->val);    
    }
    vector<int> postorderTraversal(TreeNode* root) 
    {
        vector<int> ret;
        //way-1 递归
        /*
        recursive(root,ret);
        */
         
        //way-2 非递归
        TreeNode* p = root;
        stack<TreeNode*> point;
        if (root)
            point.push(root);
         
        TreeNode* last = root;   
        while ( !point.empty()) 
        {
            p = point.top();         
            if(last == p->left || last == p->right)
            {
                ret.push_back(p->val);
                last = p;
                point.pop(); 
                continue; 
            }
              
            if (p->right)
                point.push(p->right);
            if (p->left)
                point.push(p->left);
              
            if ( !p->right && !p->left )
            {
                ret.push_back(p->val);
                last = p;
                point.pop();
            }         
        } 
        return ret;
    }
};