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?
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1 / 2 3 / 4 5The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}"
.使用非递归的方法中序遍历二叉树。
使用一个栈来模拟递归过程,如果游走指针不为NULL,就把节点入栈,继续访问其左孩子。否则的话访问栈顶元素,并把指针指向他的右孩子。访问栈顶和跳转到右孩子一定要在一个语句块中实现,原因有二,其一为从栈顶弹出的指针的左孩子一定已经访问过了,所以按中序遍历的顺序应该访问该节点,其二这样做就无需判断游走指针指向的节点的左孩子是否访问过,应为栈顶的元素的左孩子一定访问过,只需访问其右孩子即可。
代码如下:
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 11 class Solution { 12 public: 13 vector<int> inorderTraversal(TreeNode *root) { 14 vector<int> res; 15 if( root == NULL ) 16 { 17 return res; 18 } 19 TreeNode * tr = root; 20 21 stack<TreeNode*> s; 22 while(s.size()>0 || tr != NULL) 23 { 24 if(tr!=NULL) 25 { 26 s.push(tr); 27 tr= tr->left; 28 } 29 else 30 { 31 tr = s.top(); 32 s.pop(); 33 res.push_back(tr->val); 34 tr = tr->right; 35 } 36 } 37 return res; 38 } 39 };
并附上其递归版本:
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 void T(TreeNode * root,vector<int> & r) 13 { 14 if(root==NULL) return; 15 T(root->left,r); 16 r.push_back(root->val); 17 T(root->right,r); 18 } 19 vector<int> inorderTraversal(TreeNode *root) { 20 // Start typing your C/C++ solution below 21 // DO NOT write int main() function 22 vector<int> r; 23 T(root,r); 24 return r; 25 } 26 };