二叉树经典例题

发布时间：2023年12月31日

94 中序遍历

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

104 二叉树的最大深度

左右树的最大深度 + 1

/**
 * 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:
    int maxDepth(TreeNode* root) {
        if (root == nullptr) return 0;
        return max(maxDepth(root->left), maxDepth(root->right)) + 1;
    }
};

111 最小深度

/**
 * 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:
    int minDepth(TreeNode* root) {
        if (root == nullptr) return 0;
        queue<TreeNode*> q;
        q.push(root);
        int depth = 1;

        while (!q.empty()) {
            int sz = q.size();
            for (int i = 0; i < sz; ++i) {
                auto cur = q.front();
                q.pop();
                if (cur->left == nullptr && cur->right == nullptr) return depth;
                if (cur->left != nullptr) q.push(cur->left);
                if (cur->right != nullptr) q.push(cur->right);
            }
            depth++;
        }
        return depth;

    }
};

226 翻转二叉树

/**
 * 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;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        TreeNode* tmp = root->left;
        root->left = root->right;
        root->right = tmp;
        dfs(root->left);
        dfs(root->right);
    }
};

101 对称二叉树

/**
 * 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 check(TreeNode* p1, TreeNode* p2) {
        if (!p1 && !p2) {
            return true;
        }
        if (!p1 || !p2) {
            return false;
        }

        return (p1->val == p2->val) && (check(p1->left, p2->right)) && (check(p1->right, p2->left));
    }
    bool isSymmetric(TreeNode* root) {
        return check(root, root);
    }
};

222 完全二叉树节点数量（从左到右，从上到下依次排满）

复杂度 O(N)

/**
 * 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:
    int count = 0; // 统计节点数
    void inorder(TreeNode* root) {
        if(!root) return;
        inorder(root->left);
        count += 1;
        inorder(root->right);

    }
    int countNodes(TreeNode* root) {
        inorder(root);
        return count;
    }
};

2k 2k+1

h层，则最底层最右边元素是2^{h-1}最左边是2^{h}-1

二分

257 二叉树的所有路径

/**
 * 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:
    vector<string> binaryTreePaths(TreeNode* root) {
        // 创建一个存储路径的结果集
        vector<string> result;
        // 开始深度优先搜索
        dfs(root, "", result);
        return result;
    }

    void dfs(TreeNode* node, string path, vector<string>& result) {
        // 如果节点为空，返回
        if (node == nullptr) {
            return;
        }

        // *将当前节点值添加到路径中
        path += to_string(node->val);

        // 如果是叶子节点，将路径添加到结果集中
        if (node->left == nullptr && node->right == nullptr) {
            result.push_back(path);
            return;
        }

        // 如果不是叶子节点，继续深度优先搜索
        path += "->";
        dfs(node->left, path, result);
        dfs(node->right, path, result);
    }
};

404 左叶子之和

/**
 * 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:
    int ans = 0;

    int sumOfLeftLeaves(TreeNode* root) {
        dfs(root);
        return ans;
    }

    void dfs(TreeNode* r) {
        if (r == nullptr) return;
        // 是左节点且是叶子节点
        if (r->left != nullptr && r->left->left == nullptr && r->left->right == nullptr) {
            ans += r->left->val;
        }
        dfs(r->left);
        dfs(r->right);
    }
};

100 相同的树

/**
 * 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 isSameTree(TreeNode* p, TreeNode* q) {
        return dfs(p, q);
    }

    bool dfs(TreeNode* r1, TreeNode* r2) {
        if (r1 == nullptr || r2 == nullptr) return r1 == r2;
        if (r1->val != r2->val) return false;
        return dfs(r1->left, r2->left) && dfs(r1->right, r2->right);
    }
};

513 树最左下角的值

/**
 * 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:
    int findBottomLeftValue(TreeNode* root) {
        // 更新每一层的第一个节点即可
        queue<TreeNode*> q;
        q.push(root);   
        int res = root->val;

        while (!q.empty()) {
            int cs = q.size();
            for (int i = 0; i < cs; ++i) {
                auto node = q.front();
                q.pop();
                if (i == 0) res = node->val;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return res;
    }
};

543 二叉树的直径

class Solution {
    int ans;
    int depth(TreeNode* rt){
        if (rt == NULL) {
            return 0; // 访问到空节点了，返回0
        }
        int L = depth(rt->left); // 左儿子为根的子树的深度（即节点数）
        int R = depth(rt->right); // 右儿子为根的子树的深度
        ans = max(ans, L + R + 1); // 计算d_node即L+R+1 并更新ans
        return max(L, R) + 1; // 返回该节点为根的子树的深度
    }
public:
    int diameterOfBinaryTree(TreeNode* root) {
        ans = 1;
        depth(root);
        return ans - 1;
    }
};

102 二叉树的层序遍历

/**
 * 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:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;

        if (!root) {
            return res;
        }

        queue<TreeNode*> q;
        q.push(root);

        while(!q.empty()) {
            res.push_back(vector<int>());
            int currSize = q.size();

            for (int i = 1; i <= currSize; ++i) {
                auto node = q.front();
                q.pop();
                // 注意
                res.back().push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }

        }
        return res;

    }
};

110 平衡二叉树

任意节点左右子树高度差<=1

/**
 * 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:
    int height(TreeNode* root) {
        if (root == NULL) {
            return 0;
        }else {
            return max(height(root->left), height(root->right)) + 1;
        }
    }
    bool isBalanced(TreeNode* root) {
        if (root == NULL) {
            return true;
        } else {
            return abs(height(root->left) - height(root->right)) <= 1 && isBalanced(root->left) && isBalanced(root->right);
        }
    }
};

/**
 * 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 isBalanced(TreeNode* root) {
        dfs(root);
        return res;
    }

    bool res = true;
    int dfs(TreeNode* node) {
        if (node == nullptr) return 0;
        int l = dfs(node->left), r = dfs(node->right);
        if (abs(r - l) > 1) res = false;
        return std::max(l, r) + 1;
    }
};

230 二叉搜索树中第 k 小的元素

/**
 * 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:
    int kthSmallest(TreeNode* root, int k) {
        stack<TreeNode*> stack;
        while (root != nullptr || stack.size() > 0) {
            // 先找小的
            while (root != nullptr) {
                stack.push(root);
                root = root->left;
            }

            root = stack.top();
            stack.pop();
            if (--k == 0) break;
            root = root->right;
        }
        return root->val;
    }
};

建树

后序最后一个节点是根节点-->确定左子树和右子树

106 中序 + 后序遍历构建二叉树

# Definition for a binary tree node.


class Solution:
    def buildTree(self, inorder, postorder):
        # 使用哈希表存储中序遍历数组的值和对应的索引
        map = {val: idx for idx, val in enumerate(inorder)}
        # 开始递归构建树
        return self.dfs(0, len(inorder) - 1, 0, len(postorder) - 1, inorder, postorder, map)

    def dfs(self, in_left, in_right, post_left, post_right, inorder, postorder, map):
        # 如果中序遍历的左边界大于右边界，返回空节点
        if in_left > in_right:
            return None

        # 创建根节点，值为后序遍历的最后一个元素
        node = TreeNode(postorder[post_right])
        # 获取根节点在中序遍历中的索引
        in_index = map[postorder[post_right]]
        # 计算左子树节点个数
        left_cnt = in_index - in_left
        # 递归构建左右子树
        node.left = self.dfs(in_left, in_index - 1, post_left, post_left + left_cnt - 1, inorder, postorder, map)
        node.right = self.dfs(in_index + 1, in_right, post_left + left_cnt, post_right - 1, inorder, postorder, map)
        # 返回根节点
        return node

文章来源:https://blog.csdn.net/Algo_x/article/details/135279616
本文来自互联网用户投稿，该文观点仅代表作者本人，不代表本站立场。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如若内容造成侵权/违法违规/事实不符，请联系我的编程经验分享网邮箱：chenni525@qq.com进行投诉反馈，一经查实，立即删除！

二叉树 经典例题