二叉树的遍历

94. Binary Tree Inorder Traversal.cpp

class Solution {
public:
    void inorderTraversalHelper(TreeNode* root, vector<int>& x) {
        if (!root) return;
        if (root->left) inorderTraversalHelper(root->left,x);
        x.push_back(root->val);
        if (root->right) inorderTraversalHelper(root->right,x);
    }
    vector<int> inorderTraversal(TreeNode* root){
        vector<int> x;
        inorderTraversalHelper(root,x);
        return x;
    }
};

Inorder-Stack-ChangedTree-Way.cpp

/**
 * 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 {
public:
    vector<int> inorderTraversal(TreeNode* root){
        vector<int> x;
        stack<TreeNode*> s;
        if (root) s.push(root);
        TreeNode* current;
        
        while (!s.empty()){
            current = s.top();
            if (current->left){
                s.push(current->left);
                current->left = NULL;
            }
            else {
                x.push_back(s.top()->val);
                s.pop();
                if (current->right) s.push(current->right);
            }
        }
        return x;
    }
};

Inorder-Stack-Way.cpp

/*
先找到最最左的点，然后每次回溯到一个节点A的时候，都保证这个节点A的左子树已经遍历，后面就只需要遍历A的右子树(同理，又是先找到右子树的最左点)
*/
/**
 * 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 {
public:
    vector<int> inorderTraversal(TreeNode* root){
        vector<int> x;
        stack<TreeNode*> s;
        TreeNode* current=root;
        
        while (!s.empty() || current){
            if (current != NULL){
                s.push(current);
                current = current->left;
            }
            else {
                TreeNode* pNode = s.top();
                x.push_back(pNode->val);
                s.pop();
                current = pNode->right;
            }
        }
        return x;
    }
};

我的naive method，使用了递归。没什么好说的。
Morris Traversal 竟然可以不要Stack.我也是醉了。