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LeetCode Binary Tree Inorder Traversal

时间:2014-03-23 10:24:53      阅读:375      评论:0      收藏:0      [点我收藏+]
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class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> res;
        if (root == NULL) return res;
        vector<pair<TreeNode*, bool> > stack;
        stack.push_back(make_pair(root, false));
        
        while (!stack.empty()) {
            pair<TreeNode*, bool>& pn = stack.back();
            TreeNode* n = pn.first;
            if (pn.second) {
                stack.pop_back();
                res.push_back(n->val);
                if (n->right != NULL) {
                    stack.push_back(make_pair(n->right, false));
                }
                continue;
            }
            pn.second = true;
            n = n->left;
            while (n != NULL) {
                stack.push_back(make_pair(n, true));
                n = n->left;
            }
        }
    }
    
    // a better one
    vector<int> _inorderTraversal(TreeNode *root) {
        vector<int> res;
        vector<TreeNode*> stack;
        
        while(root != NULL || !stack.empty()) {
                while (root != NULL) {
                    stack.push_back(root);    
                    root=root->left;
                }
                if (!stack.empty()) {
                    root = stack.back();
                    stack.pop_back();
                    res.push_back(root->val); 
                    root = root->right;   
                }
        }
    }
};
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这种题想想简单,写起来也不好写,尤其是写的像第二个那么巧妙简洁,又找到一个比较通用的可以用在中序遍历和后续遍历中

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class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
                vector<int> res;
        if (root == NULL) return res;
        vector<pair<TreeNode*, bool> > stack;
        stack.push_back(make_pair(root, false));
        
        while (!stack.empty()) {
            pair<TreeNode*, bool>& pn = stack.back();
            TreeNode* n = pn.first;
            stack.pop_back();
            
            if (pn.second) {
                res.push_back(n->val);
            } else {
                if (n->right != NULL) {
                    stack.push_back(make_pair(n->right, false));
                }
                stack.push_back(make_pair(n, true));
                if (n->left != NULL) {
                    stack.push_back(make_pair(n->left, false));    
                }
            }
        }
        return res;
    }
};
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LeetCode Binary Tree Inorder Traversal

原文:http://www.cnblogs.com/lailailai/p/3615605.html

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