144.Binary Tree Preorder Traversal

Tags: Medium Tree

Link: https://leetcode.com/problems/binary-tree-preorder-traversal/

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,2,3]

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

---

Answer:

递归写法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if (!root) return res;

        preorderTraversal(root, res);

        return res;
    }

    void preorderTraversal(TreeNode *p, vector<int> & res) {
        if (!p) return;

        res.push_back(p -> val);
        if (p -> left) preorderTraversal(p -> left, res);
        if (p -> right) preorderTraversal(p -> right, res);
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Binary Tree Preorder Traversal.
Memory Usage: 9.5 MB, less than 17.24% of C++ online submissions for Binary Tree Preorder Traversal.

使用栈的写法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if (!root) return res;

        stack<TreeNode *> store;
        store.push(root);

        while (!store.empty()) {
            TreeNode *p = store.top();
            store.pop();
            res.push_back(p -> val);
            if (p -> right) store.push(p -> right);
            if (p -> left) store.push(p -> left);
        }

        return res;
    }
};

Runtime: 4 ms, faster than 59.43% of C++ online submissions for Binary Tree Preorder Traversal.
Memory Usage: 9 MB, less than 100.00% of C++ online submissions for Binary Tree Preorder Traversal.

Morris遍历（线索二叉树）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        if (!root) return res;

        TreeNode * cur = root, *pre = nullptr;
        while (cur != nullptr) {
            if (cur -> left == nullptr) {
                res.push_back(cur -> val);
                cur = cur -> right;
            }
            else {
                pre = cur -> left;
                while (pre -> right != nullptr && pre -> right != cur)
                    pre = pre -> right; //当前节点左子树的最右子节点

                if (pre -> right == nullptr) {
                    pre -> right = cur;
                    res.push_back(cur -> val);
                    cur = cur -> left;
                }
                else {
                    pre -> right = nullptr;
                    cur = cur -> right;
                }
            }
        }

        return res;
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Binary Tree Preorder Traversal.
Memory Usage: 8.9 MB, less than 100.00% of C++ online submissions for Binary Tree Preorder Traversal.

• 如果当前节点的左孩子为空，则输出当前节点并将其右孩子作为当前节点。
• 如果当前节点的左孩子不为空，在当前节点的左子树中找到当前节点在中序遍历下的前驱节点。
• 如果前驱节点的右孩子为空，将它的右孩子设置为当前节点。**输出当前节点（在这里输出，这是与中序遍历唯一一点不同）。**当前节点更新为当前节点的左孩子。
• 如果前驱节点的右孩子为当前节点，将它的右孩子重新设为空。当前节点更新为当前节点的右孩子。
• 重复以上1、2直到当前节点为空。