day15 | 二叉树2

0.引言

● 层序遍历 10
● 226.翻转二叉树
● 101.对称二叉树 2

1.翻转二叉树

给你一棵二叉树的根节点 root ，翻转这棵二叉树，并返回其根节点。

示例 1：

image.png

输入：root = [4,2,7,1,3,6,9]
输出：[4,7,2,9,6,3,1]

示例 2：

image.png

输入：root = [2,1,3]
输出：[2,3,1]

示例 3：

输入：root = []
输出：[]

提示：

• 树中节点数目范围在 [0, 100] 内
• -100 <= Node.val <= 100

Discussion | Solution

1.1递归法

/*
 * @lc app=leetcode.cn id=226 lang=cpp
 *
 * [226] 翻转二叉树
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
 public:
  TreeNode* invertTree(TreeNode* root) {
    dfs(root);
    return root;
  }
  // 前序遍历
 private:
  void dfs(TreeNode* node) {
    if (node == nullptr) return;
    std::swap(cur->left, cur->right);
    dfs(node->left);
    dfs(node->right);
  }
};
// @lc code=end

1.2.迭代法

/*
 * @lc app=leetcode.cn id=226 lang=cpp
 *
 * [226] 翻转二叉树
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
 public:
  TreeNode* invertTree(TreeNode* root) {
    std::stack<TreeNode*> node_stack;
    if (root == nullptr) return root;
    node_stack.push(root);

    while (!node_stack.empty()) {
      TreeNode* cur = node_stack.top();
      node_stack.pop();
      std::swap(cur->left, cur->right);
      if (cur->left) node_stack.push(cur->left);
      if (cur->right) node_stack.push(cur->right);
    }
    return root;
  }
};
// @lc code=end

1.3.层序遍历BFS

/*
 * @lc app=leetcode.cn id=226 lang=cpp
 *
 * [226] 翻转二叉树
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
 public:
  TreeNode* invertTree(TreeNode* root) {
    std::queue<TreeNode*> que;
    if (root != nullptr) {
      que.push(root);
    }
    while (!que.empty()) {
      for (int i = que.size(); i > 0; --i) {
        TreeNode* cur = que.front();
        que.pop();
        std::swap(cur->left, cur->right);
        if (cur->left) que.push(cur->left);
        if (cur->right) que.push(cur->right);
      }
    }
    return root;
  }
};
// @lc code=end

2.对称二叉树

给你一个二叉树的根节点 root ， 检查它是否轴对称。

示例 1：

image.png

输入：root = [1,2,2,3,4,4,3]
输出：true

示例 2：

image.png

输入：root = [1,2,2,null,3,null,3]
输出：false

提示：

• 树中节点数目在范围 [1, 1000] 内
• -100 <= Node.val <= 100

进阶：你可以运用递归和迭代两种方法解决这个问题吗？

Discussion | Solution

2.1.递归法

/*
 * @lc app=leetcode.cn id=101 lang=cpp
 *
 * [101] 对称二叉树
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
 public:
  bool isSymmetric(TreeNode* root) { return dfs(root, root); }
  // 后续遍历
 private:
  bool dfs(TreeNode* left, TreeNode* right) {
    if (left == nullptr && right == nullptr) return true;
    if (left == nullptr || right == nullptr) return false;
    if (left->val != right->val) return false;

    bool out_side = dfs(left->left, right->right);
    bool in_side = dfs(left->right, right->left);
    return in_side && out_side;
  }
};
// @lc code=end

2.2.迭代法-队列

/*
 * @lc app=leetcode.cn id=101 lang=cpp
 *
 * [101] 对称二叉树
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
 public:
  bool isSymmetric(TreeNode* root) {
    if (root == nullptr) return true;
    std::queue<TreeNode*> que;
    que.push(root->left);
    que.push(root->right);

    while (!que.empty()) {
      TreeNode* left = que.front();
      que.pop();
      TreeNode* right = que.front();
      que.pop();

      // 判断逻辑，与递归逻辑相同
      if (left == nullptr && right == nullptr) continue;
      if (left == nullptr || right == nullptr) return false;
      if (left->val != right->val) return false;

      // 外侧
      que.push(left->left);
      que.push(right->right);
      // 内侧
      que.push(left->right);
      que.push(right->left);
    }
    return true;
  }
};
// @lc code=end

2.3.迭代法-栈

/*
 * @lc app=leetcode.cn id=101 lang=cpp
 *
 * [101] 对称二叉树
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
 public:
  bool isSymmetric(TreeNode* root) {
    if (root == nullptr) return true;
    std::stack<TreeNode*> st;
    st.push(root->left);
    st.push(root->right);

    while (!st.empty()) {
      TreeNode* left = st.top();
      st.pop();
      TreeNode* right = st.top();
      st.pop();

      // 判断逻辑，与递归逻辑相同
      if (left == nullptr && right == nullptr) continue;
      if (left == nullptr || right == nullptr) return false;
      if (left->val != right->val) return false;

      // 外侧
      st.push(left->left);
      st.push(right->right);
      // 内侧
      st.push(left->right);
      st.push(right->left);
    }
    return true;
  }
};
// @lc code=end