• leetcode[94]Binary Tree Inorder Traversal


    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.

    /**
     * Definition for binary tree
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
    vector<int> inorderTraversal(TreeNode *root) 
    {
        vector<int> res;
        if(root==NULL)return res;
        TreeNode *p=root;
        stack<TreeNode *> sta;
        while(p!=NULL||!sta.empty())
        {
            while(p!=NULL)
            {
                sta.push(p);
                p=p->left;
            }
            if(!sta.empty())
            {
                p=sta.top();
                sta.pop();
                res.push_back(p->val);
                p=p->right;
            }
        }
        return res;
    }
    /*
    void inorder(TreeNode *root, vector<int> &res)
    {
        if(root==NULL)return;
        inorder(root->left,res);
        res.push_back(root->val);
        inorder(root->right,res);
        return;
    }
        vector<int> inorderTraversal(TreeNode *root) {
            vector<int> res;
            inorder(root,res);
            return res;
        }
    */
    };
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  • 原文地址:https://www.cnblogs.com/Vae1990Silence/p/4281382.html
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