LeetCode 94.Binary Tree Inorder Traversal (二叉树的中序遍历)

Bertha 。 2022-05-09 04:54 126阅读 0赞

题目描述：

给定一个二叉树，返回它的*中序* 遍历。

**示例:**

输入: [1,null,2,3]
       1
        \
         2
        /
       3
    
    输出: [1,3,2]

**进阶:** 递归算法很简单，你可以通过迭代算法完成吗？

AC C++ Solution:  
1.递归法：

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    
    /* 递归法 */
    class Solution {
    public:
        vector<int> inorderTraversal(TreeNode* root) {
            vector<int> nodes;
            inorder(root,nodes);
            return nodes;
        }
        
    private:
        void inorder(TreeNode* root, vector<int>& nodes) {
            if(!root)
                return;
            
            inorder(root->left,nodes);  //当左子节点为空时,访问当前节点值
            nodes.push_back(root->val);
            inorder(root->right,nodes);
        }
    };

2.堆栈法：

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    
    /* 堆栈法*/
    class Solution {
    public:
        vector<int> inorderTraversal(TreeNode* root) {
            vector<int> vector;
            stack<TreeNode *> stack;
            TreeNode *pCurrent = root;
            
            while(!stack.empty() || pCurrent)
            {
               if(pCurrent) {
                   stack.push(pCurrent);
                   pCurrent =  pCurrent->left;
               }
                else {
                    pCurrent = stack.top();
                    stack.pop();
                    vector.push_back(pCurrent->val);
                    pCurrent = pCurrent->right;
                }
            }
            return vector;
        }
    };
    };