[leetCode]Binary Tree Inorder Traversal 递归 && 栈解法

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

#include <vector>
#include <iostream>
#include <stack>
using namespace std;
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    vector<int> valList;
    vector<int> inorderTraversal1(TreeNode *root) {
        InorderWalk(root);
        return valList;
    }
    void  InorderWalk(TreeNode *root){
        if(root == NULL) return;
        TreeNode *cur = root;
        if(cur->left != NULL) InorderWalk(cur->left);
        valList.push_back(cur->val);
        if(cur->right != NULL) InorderWalk(cur->right);        
    }
}

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.

    /**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
#include <vector>
#include <iostream>
#include <stack>
using namespace std;
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    vector<int> valList;stack<TreeNode*> stk;
    vector<int> inorderTraversal2(TreeNode *root) {//Iteratively    
        valList.clear();
        TreeNode *cur = root;
        cur = SearchLeft(root);     
        while(!stk.empty()){
            cur = stk.top();
            stk.pop();
            valList.push_back(cur->val);
            if(cur->right != NULL){
                cur = cur->right;
                cur = SearchLeft(cur);
            }
        }
        return valList;
    }
private:
    TreeNode* SearchLeft(TreeNode *root){
        if(root == NULL) return NULL;
        TreeNode *cur = root;
        while(cur->left !=NULL || cur->right != NULL){
            if(cur->left != NULL){
                stk.push(cur);
                cur = cur->left;
                continue;
            }
            valList.push_back(cur->val);
            cur = cur->right;
        }
        valList.push_back(cur->val);
        return cur;
    }