leetcode226

最新推荐文章于 2019-10-26 10:09:32 发布

原创最新推荐文章于 2019-10-26 10:09:32 发布 · 205 阅读

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本文详细介绍了两种翻转二叉树的方法：递归解法和非递归解法。递归解法通过交换根节点的左右子树并递归处理子树实现；非递归解法使用队列遍历树，逐层交换节点的左右子树。

226. Invert Binary Tree

Description：

Invert a binary tree.

Example:

Input:

     4
   /   \
  2     7
 / \   / \
1   3 6   9

Output:

     4
   /   \
  7     2
 / \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

思路一：递归解法：交换根节点的左右子树，再分别对左右子树执行递归，终止条件为根节点没有左右子树时。

代码：

/**
 * 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:
    TreeNode* invertTree(TreeNode* root) {
        if(root == NULL) return root;
        TreeNode* tempRoot = root->left;
        root->left = invertTree(root->right);
        root->right = invertTree(tempRoot);
        return root;
    }
};

思路二：非递归解法，利用队列遍历树，从根节点开始依次交换每层节点的左右子节点。

代码：

/**
 * 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:
    TreeNode* invertTree(TreeNode* root) {
        if(root == NULL) return root;
        queue <TreeNode*> tree;
        tree.push(root);
        while(tree.size() > 0) {
            TreeNode *currentRoot = tree.front();
			tree.pop();
			TreeNode *temp = currentRoot->left;
			currentRoot->left = currentRoot->right;
			currentRoot->right = temp;
			if (currentRoot->left) {
                tree.push(currentRoot->left);
            }	
			if (currentRoot->right) {
                tree.push(currentRoot->right);
            }	
        }
        return root;
    }
};