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LeetCode: Binary Tree Inorder Traversal

LeetCode: 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?

地址:https://oj.leetcode.com/problems/binary-tree-inorder-traversal/

算法:要求用非递归来进行中序遍历。初始时,从根节点开始,一直往左走,并把每个节点入栈。在while循环内,取栈顶元素,出栈,访问该元素,若该元素有右孩子,往右走,在一直往左走到低并把每个节点进栈,如此循环知道栈为空。代码:

 1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     vector<int> inorderTraversal(TreeNode *root) {13         vector<int> result;14         if(!root)   return result;15         TreeNode *p = root;16         stack<TreeNode*> stk;17         while(p){18             stk.push(p);19             p = p->left;20         }21         while(!stk.empty()){22             p = stk.top();23             stk.pop();24             result.push_back(p->val);25             p = p->right;26             while(p){27                 stk.push(p);28                 p = p->left;29             }30         }31         return result;32     }33 };

 

LeetCode: Binary Tree Inorder Traversal