Binary Tree Inorder Traversal

标签：leetcode   binary tree inorder   

1. 递归解法

/**
 * 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> result;
    void inorder(TreeNode *root)
    {
        if (root == NULL)
            return;
        inorder(root->left);
        result.push_back(root->val);
        inorder(root->right);
    }
    vector<int> inorderTraversal(TreeNode *root) {
        result.clear();
        inorder(root);
        return result;
    }
};

2. 非递归解法（空间复杂度O(n))

/**
 * 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> path;
        stack<TreeNode *> st;
        if (root == NULL)
            return path;
        TreeNode *p = root;
        while (p != NULL || !st.empty())
        {
            while (p != NULL)
            {
                st.push(p);
                p = p->left;
            }
            if (!st.empty())
            {
                p = st.top();
                st.pop();
                path.push_back(p->val);
                p = p->right;
            }
        }
        return path;
    }
};

3. 非递归解法（空间复杂度O(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> inorderTraversal(TreeNode *root) {
        vector<int> path;
        if (root == NULL)
            return path;
            
        TreeNode *cur = root;
        TreeNode *pre = NULL;
        while (cur != NULL)
        {
            if (cur->left == NULL)
            {
                path.push_back(cur->val);
                cur = cur->right;
            }
            else
            {
                pre = cur->left;
                while (pre->right != NULL && pre->right != cur)
                    pre = pre->right;
                if (pre->right == NULL)
                {
                    pre->right = cur;
                    cur = cur->left;
                }
                else
                {
                    pre->right = NULL;
                    path.push_back(cur->val);
                    cur = cur->right;
                }
            }
        }
        return path;
    }
};

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Binary Tree Inorder Traversal

标签：leetcode   binary tree inorder   

原文地址：http://blog.csdn.net/hevc_cjl/article/details/28681977