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?
2种解法，一种是正常递归，另一种是使用栈思想
1 #include <vector> 2 #include <iostream> 3 #include <stack> 4 using namespace std; 5 struct TreeNode { 6 int val; 7 TreeNode *left; 8 TreeNode *right; 9 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 10 }; 11 class Solution { 12 public: 13 vector<int> valList; 14 vector<int> inorderTraversal1(TreeNode *root) { 15 InorderWalk(root); 16 return valList; 17 } 18 void InorderWalk(TreeNode *root){ 19 if(root == NULL) return; 20 TreeNode *cur = root; 21 if(cur->left != NULL) InorderWalk(cur->left); 22 valList.push_back(cur->val); 23 if(cur->right != NULL) InorderWalk(cur->right); 24 } 25 }
第二种，使用堆栈。比较重要的函数是 SearchLeft(TreeNode *root)，他的作用是寻找一个在左子树上的叶子节点，并将路径上出现的节点压栈，若一条路径上没有左子树则将其加入输出队列(不知道解释清楚没)
中心思想大概可以理解为：有左边就(自己压栈再)走左边，没有左边就(自己输出再)走右边，左右都没有就输出自己。
1.若当前节点有左子树或右子树，或都有则进入2.否则表示当前节点为叶子节点，将其加入输出队列(valList)。
2.若当前节点(cur)有左子树(cur->left!=NULL)，将自己压栈，并走左子树(cur = cur->left)。再次判断1.
3.若当前节点没有左子树，则将自己加入输出队列(valList)中。再判断，若当前节点有右子树，走右子树，再次判断1.
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 #include <vector> 11 #include <iostream> 12 #include <stack> 13 using namespace std; 14 struct TreeNode { 15 int val; 16 TreeNode *left; 17 TreeNode *right; 18 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 19 }; 20 class Solution { 21 public: 22 vector<int> valList;stack<TreeNode*> stk; 23 vector<int> inorderTraversal2(TreeNode *root) {//Iteratively 24 valList.clear(); 25 TreeNode *cur = root; 26 cur = SearchLeft(root); 27 while(!stk.empty()){ 28 cur = stk.top(); 29 stk.pop(); 30 valList.push_back(cur->val); 31 if(cur->right != NULL){ 32 cur = cur->right; 33 cur = SearchLeft(cur); 34 } 35 } 36 return valList; 37 } 38 private: 39 TreeNode* SearchLeft(TreeNode *root){ 40 if(root == NULL) return NULL; 41 TreeNode *cur = root; 42 while(cur->left !=NULL || cur->right != NULL){ 43 if(cur->left != NULL){ 44 stk.push(cur); 45 cur = cur->left; 46 continue; 47 } 48 valList.push_back(cur->val); 49 cur = cur->right; 50 } 51 valList.push_back(cur->val); 52 return cur; 53 }