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},
非递归版本：
用一个栈来模拟递归的情况，
首先思考inorder traverse的递归函数
traverse(left tree);
visit current node
traverse(right tree);
也就是说我们把一个节点压入栈中，首先它会先递归访问左子树（左节点入栈），再访问本身（这个时候这个节点就可以出栈了）,在访问右子树（右节点入栈）。
最后我们定义一个数据结构
1 struct Node 2 { 3 TreeNode *tNode; 4 bool findLeft; 5 Node(){} 6 Node(TreeNode *n):tNode(n), findLeft(false){} 7 };
其中findLeft是判断是否已将左节点入栈了
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 struct Node 11 { 12 TreeNode *tNode; 13 bool findLeft; 14 Node(){} 15 Node(TreeNode *n):tNode(n), findLeft(false){} 16 }; 17 18 class Solution { 19 private: 20 vector<int> ret; 21 public: 22 vector<int> inorderTraversal(TreeNode *root) { 23 // Start typing your C/C++ solution below 24 // DO NOT write int main() function 25 stack<Node> s; 26 s.push(Node(root, false)); 27 ret.clear(); 28 29 while(!s.empty()) 30 { 31 Node node = s.top(); 32 if (node.tNode == NULL) 33 s.pop(); 34 else 35 { 36 if (!node.findLeft) 37 { 38 s.pop(); 39 s.push(Node(node.tNode, true)); 40 s.push(Node(node.tNode->left, false)); 41 } 42 else 43 { 44 s.pop(); 45 ret.push_back(node.tNode->val); 46 s.push(Node(node.tNode->right, false)); 47 } 48 } 49 } 50 51 return ret; 52 } 53 };
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 private: 12 stack<TreeNode* > s; 13 vector<int> ret; 14 public: 15 vector<int> inorderTraversal(TreeNode *root) { 16 // Start typing your C/C++ solution below 17 // DO NOT write int main() function 18 ret.clear(); 19 if (root == NULL) 20 return ret; 21 22 s.push(root); 23 TreeNode *node = root; 24 while(!s.empty()) 25 { 26 node = node->left; 27 s.push(node); 28 } 29 } 30 };
