Leetcode: Binary Tree Inorder Traversal

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?

分析：迭代版inorder traversal。对于任何一棵树，其最先被访问的是路径是最左侧的路径，在该最左侧路径上访问顺序呢是从叶子结点到根节点，很自然我们可以用一个stack保存这个路径。同时如果当前访问结点有右孩子，我们需把以右孩子为根的子树最左路径压入栈中。时间复杂度为O(n)，空间复杂度为O(h)。代码如下：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> result;
        if(!root) return result;
        
        stack<TreeNode*> S;
        for(TreeNode *p = root; p; p = p->left)
            S.push(p);
        
        while(!S.empty()){
            TreeNode *tmp = S.top();
            S.pop();
            result.push_back(tmp->val);
            for(TreeNode *p = tmp->right; p ; p = p->left)
                S.push(p);
        }
        
        return result;
    }
};

 Morris中序遍历：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> result;
        TreeNode * prev = NULL, *cur = root;
        
        while(cur != NULL){
            if(cur->left == NULL){
                result.push_back(cur->val);
                cur = cur->right;
            }else{
                for(prev = cur->left; prev->right != NULL && prev->right != cur; prev = prev->right);
                if(prev->right == NULL){
                    prev->right = cur;
                    cur = cur->left;
                }else{
                    prev->right = NULL;
                    result.push_back(cur->val);
                    cur = cur->right;
                }
            }
        }
        
        return result;
    }
};