LC 144. / 94. / 145. Binary Tree Preorder/ Inorder/ PostOrder Traversal

题目描述

144. Binary Tree Preorder Traversal

94. Binary Tree Inorder Traversal

145. Binary Tree Postorder Traversal

前序排列 :根-左-右

中序排列: 左-根-右

后序排列:左-右-根

参考答案

 // PreOrder
 /**
  * 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 {
 private:
         vector<int> res;

 public:
     vector<int> preorderTraversal(TreeNode* root) {
         foo(root);
         return res;
     }
     void foo(TreeNode* root){
         if(root == NULL)
             return;

         res.push_back(root->val);
         foo(root->left);
         foo(root->right);
     }
 };

 // InOrder
 class Solution {
 public:
     vector<int> inorderTraversal(TreeNode* root) {
         if(root == NULL){
             return vector<int>();
         }
         vector<int> result; // 创建结果
         stack<TreeNode *> s;// 创建堆栈
         TreeNode *p = root; // 临时让p为root
         // 寻找p最左边的node
         while(p!=NULL){
             s.push(p); // 从root开始，将左边的node推入stack
             p = p->left;// 更新p为左node
         }
         // s 为全是左节点的stack
         // 对 s进行循环操作
         while(!s.empty()){
             // 将最最左边的，推入stack
             p = s.top();
             result.push_back(p->val);
             s.pop(); // 自己消失了
             // 然后观察这个的右边node
             if(p->right != NULL){
                 p = p->right;
                 while(p!=NULL){ //观察node的左边
                     s.push(p);
                     p = p->left;
                 }
             }
         }
         return result;
     }
 };

 // PostOrder

 class Solution {
 public:
     vector<int> postorderTraversal(TreeNode* root) {
         if(root == NULL){
             return vector<int>();
         }
         vector<int> result;
         stack<TreeNode *> s;

         s.push(root);
         while(!s.empty()){
             TreeNode *temp = s.top();
             result.push_back(temp->val);
             s.pop();
             if(temp->left!=NULL){
                 s.push(temp->left);
             }
             if(temp->right!= NULL){
                 s.push(temp->right);
             }
         }
         reverse(result.begin(), result.end());
         return result;

     }
 };