[leetcode 94] Binary Tree Inorder Traversal

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本文介绍三种实现二叉树中序遍历的方法：递归、迭代使用栈及Morris遍历。递归方法直观易懂，但空间复杂度较高；迭代方法利用栈结构进行节点管理，避免了递归调用；Morris遍历则是一种巧妙地利用空闲指针进行遍历的方法，空间复杂度为O(1)。

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

递归

/**
 * 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> res;
    vector<int> inorderTraversal(TreeNode *root) {
        if (!root) {
            return res;
        }
        inorderTraversal(root->left);
        res.push_back(root->val);
        inorderTraversal(root->right);
        return res;
    }
};

栈

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

    }
};

Morris 遍历

/**
 * 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> res;
        if (!root) {
            return res;
        }
        TreeNode *cur = root;
        while (cur) {
            if (cur->left == NULL) {
                res.push_back(cur->val);
                cur = cur->right;
            } else {
                auto node = cur->left;
                while (node->right && node->right != cur) {
                    node = node->right;
                }
                if (!node->right) {
                    node->right = cur;
                    cur = cur->left;
                } else {
                    res.push_back(cur->val);
                    node->right = NULL;
                    cur = cur->right;
                }
            }
        }
        
        
        return res;

    }
};