[leetcode]binary-tree-inorder-traversal

思路

1. 递归
2. 非递归-回溯

---

Solution 1

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> outputVec;
    vector<int> inorderTraversal(TreeNode *root) {
        if(root!=nullptr){
            inorderTraversal(root->left);
            outputVec.push_back(root->val);
            inorderTraversal(root->right);
        }
        return outputVec;
    }
};

---

Solution 2

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
          vector<int> output;
        if(!root){
            return output;
        }

        vector<TreeNode*> s;
        s.push_back(root);
        TreeNode* currNode;
        TreeNode* lastPopedNode;
        while(!s.empty()){
            currNode = s.back();
            //leaf node, just pop
            if(!currNode->left && !currNode->right){
                //output.push_back(currNode->val);//post
                output.push_back(currNode->val);//mid
                //output.push_back(currNode->val);//pre
                lastPopedNode = s.back();
                s.pop_back();
                continue;
            }

            //from left. check right
            if(lastPopedNode == currNode->left){
                output.push_back(currNode->val);//mid
                if(currNode->right){
                    s.push_back(currNode->right);
                }else{
                    //output.push_back(currNode->val);//post
                    lastPopedNode = s.back();
                    s.pop_back();
                }
                continue;
            }

            //from right, just pop
            if(lastPopedNode == currNode->right){
                //output.push_back(currNode->val);//post
                lastPopedNode = s.back();
                s.pop_back();
                continue;
            }

            //this node is pushed into stack just now 
            if(currNode->left){
                //output.push_back(currNode->val);//pre
                s.push_back(currNode->left);
                continue;
            }

            //no left , but has right
            if(currNode->right){
                //output.push_back(currNode->val);//pre
                output.push_back(currNode->val);//mid
                s.push_back(currNode->right);
                continue;
            }

        }
    }
};