leetcodeBinary Tree Inorder Traversal

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题目描述

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

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

   1    \     2    /   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

解题方法

中序遍历二叉树：递归方法：

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

迭代方法：

仿照前序遍历使用堆栈，不过是左节点先入栈，直到左孩子为空，出栈，入栈顶元素的右孩子，然后再入右子树的左孩子节点。依次直到堆栈为空，值得注意的是，中序遍历跟先序遍历起始不同的是，根节点不最先入栈！！！代码如下：

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

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