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[Tree]Binary Tree Inorder Traversal

时间:2015-12-12 01:38:25      阅读:232      评论:0      收藏:0      [点我收藏+]
Total Accepted: 98729 Total Submissions: 261539 Difficulty: Medium

 

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?

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

[Tree]Binary Tree Inorder Traversal

原文:http://www.cnblogs.com/zengzy/p/5040640.html

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