【Leetcode】Binary Tree Traversal

把三个二叉树遍历的题放在一起了。

递归写法太简单，就不再实现了，每题实现了两种非递归算法。

一种是利用栈，时间和空间复杂度都是O(n)。

另一种是借助线索二叉树，也叫Morris遍历，充分利用树中节点的空指针域。

先序：

Binary Tree Preorder Traversal

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

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

   1
    \
     2
    /
   3

return [1,2,3].

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

class Solution {
public:
    vector<int> preorderTraversal(TreeNode *root) {
        vector<int> result;
        TreeNode *p = root;
        while (p) {
            if (!p->left) {
                result.push_back(p->val);
                p = p->right;
            } else {
                TreeNode *q = p->left;
                while (q->right && q->right != p) {
                    q = q->right;
                }
                if (q->right == nullptr) {
                    result.push_back(p->val);
                    q->right = p;
                    p = p->left;
                } else {
                    q->right = nullptr;
                    p = p->right;
                }
            }
        }
        return result;
    }
};

---

中序：

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].

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

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> result;
        TreeNode *p = root;
        while (p) {
            if (p->left) {
                TreeNode *q = p->left;
                while (q->right && q->right != p) {
                    q = q->right;
                }
                if (q->right == p) {
                    result.push_back(p->val);
                    q->right = nullptr;
                    p = p->right;
                } else {
                    q->right = p;
                    p = p->left;
                }
            } else {
                result.push_back(p->val);
                p = p->right;
            }
        }
        return result;
    }
};

---

后序：

Binary Tree Postorder Traversal

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

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

   1
    \
     2
    /
   3

return [3,2,1].

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> result;
        stack<TreeNode *> s;
        TreeNode *p = root;
        TreeNode *q = nullptr, *last = nullptr;
        while (!s.empty() || p) {
            if (p) {
                s.push(p);
                p = p->left;
            } else {
                q = s.top();
                if (q->right && last != q->right) {
                    p = q->right;
                } else {
                    result.push_back(q->val);
                    s.pop();
                    last = q;
                }    
            }
        }
        return result;
    }
};

class Solution {
public:
    vector<int> postorderTraversal(TreeNode *root) {
        vector<int> result;
        TreeNode dummy(0);
        TreeNode *p = &dummy, *pre = nullptr;
        p->left = root;
        while (p) {
            if (p->left == nullptr) {
                p = p->right;
            } else {
                pre = p->left;
                while (pre->right != nullptr && pre->right != p) {
                    pre = pre->right;
                }
                if (pre->right == nullptr) {
                    pre->right = p;
                    p = p->left;
                } else {
                    printReverse(result, p->left, pre);
                    pre->right = nullptr;
                    p = p->right;
                }
            }
        }
        return result;
    }

private:
    void printReverse(vector<int>& v, TreeNode* from, TreeNode *to) {
        reverse(from, to);
        TreeNode *p = to;
        while (true) {
            v.push_back(p->val);
            if (p == from) {
                break;
            }
            p = p->right;
        }
        reverse(to, from);
    }
    
    void reverse(TreeNode *from, TreeNode *to) {
        if (from == to) {
            return;
        }
        TreeNode *x = from, *y = x->right, *z;
        while (true) {
            z = y->right;
            y->right = x;
            x = y;
            y = z;
            if (x == to) {
                break;
            }
        }
    }
};