二叉树题目-leetcode

最新推荐文章于 2025-07-16 16:35:13 发布

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最新推荐文章于 2025-07-16 16:35:13 发布

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分类专栏：刷题文章标签：刷题

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20 篇文章

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本文详细解析了多种二叉树算法问题，包括对称二叉树的判断、填充每个节点的下一个右侧节点、求根到叶的数字之和、获取二叉树右视图以及寻找二叉树中两节点的最近公共祖先。

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101 Symmetric Tree

给定一颗二叉树，检查是否镜像对称（围绕中心对称）

分析：递归，从根节点开始，判断左节点的左子树与右节点的右子树，左节点的右子树与右节点的左子树是否相等即可！

class Solution {
public:
    bool judge(TreeNode* left, TreeNode* right){
        if ((!left) && (!right)) return true;
        if ((!left) || (!right)) return false;
        if (left->val != right->val) return false;
        return judge(left->right, right->left)&&judge(left->left, right->right);
    }
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        return judge(root->left, root->right);
    }
};

非递归的写法：中间的时候不能return true，应该continue

/**
 * 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:
    bool isSymmetric(TreeNode* root) {
        if (!root) return true;
        queue <TreeNode*> que;
        que.push(root->left);
        que.push(root->right);
        TreeNode* leftNode;
        TreeNode* rightNode;
        while(!que.empty()){
            leftNode = que.front();
            que.pop();
            rightNode = que.front();
            que.pop();
            if ((!leftNode) && (!rightNode)) continue;
                //return true;是不对的
            if ((!leftNode) || (!rightNode)) return false;
            if (leftNode->val != rightNode->val) return false;
            que.push(leftNode->right);
            que.push(rightNode->left);
            que.push(leftNode->left);
            que.push(rightNode->right);
    }
        return true;
}
};

116 Populating Next Right Pointers in Each Node I

分析：层序遍历的思路，注意利用next指针，使得空间复杂度控制在O(1)。

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        if (root == NULL) return;
        root -> next = NULL;
        TreeLinkNode *curNode = root;
        while (curNode) {
             TreeLinkNode *node = curNode;
             if (!node -> left) return; 
             while (node) {
                 node -> left -> next = node -> right;
                 if(node -> next) node -> right -> next = node -> next -> left;
                 node = node -> next;
             }
             curNode = curNode -> left;
        }
    }
};

117 Populating Next Right Pointers in Each Node II

思路：层序遍历,O(1)空间复杂度和O(n)时间复杂度

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        // current level node
        TreeLinkNode *curNode = root;
        while (curNode) {
             //the first node of next level           
             TreeLinkNode *firstNode = new TreeLinkNode(0);  
             TreeLinkNode *nextNode = firstNode;
             //level-order traversal 
             while(curNode){  
                 if (curNode -> left) {
                 nextNode->next = curNode->left;
                 nextNode = nextNode->next;
                }
                if (curNode -> right) {
                 nextNode->next = curNode->right;
                 nextNode = nextNode->next;
                }  
                curNode = curNode->next; 
             }
             //jump to the next level
             curNode = firstNode->next;
             delete firstNode;
    }
    }
};

类似的思路：

public class Solution {

    //based on level order traversal
    public void connect(TreeLinkNode root) {

        TreeLinkNode head = null; //head of the next level
        TreeLinkNode prev = null; //the leading node on the next level
        TreeLinkNode cur = root;  //current node of current level

        while (cur != null) {

            while (cur != null) { //iterate on the current level
                //left child
                if (cur.left != null) {
                    if (prev != null) {
                        prev.next = cur.left;
                    } else {
                        head = cur.left;
                    }
                    prev = cur.left;
                }
                //right child
                if (cur.right != null) {
                    if (prev != null) {
                        prev.next = cur.right;
                    } else {
                        head = cur.right;
                    }
                    prev = cur.right;
                }
                //move to next node
                cur = cur.next;
            }

            //move to next level
            cur = head;
            head = null;
            prev = null;
        }

    }
}

129. Sum Root to Leaf Numbers

分析：深度优先遍历，累计当前结点的值。

/**
 * 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:
    int sumNumbers(TreeNode* root) {
        return (root, 0);
    }
    int df(TreeNode* root, int sum) {
        if (!root) return 0;
        if (!root->left&&!root->right) return sum*10+root->val;
        return df(root->left, sum*10+root->val) + df(root->right, sum*10+root->val);

    }
};

199. Binary Tree Right Side View

深度遍历或者广度遍历
广度优先遍历

/**
 * 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> rightSideView(TreeNode* root) {
    vector<int> res;
    if (!root) return res;
    queue <TreeNode*> nodes;
    nodes.push(root);
    while (!nodes.empty()) {
        res.push_back(nodes.front()->val);
        int len = nodes.size();
        for (int i = 0; i < len; ++i){
            TreeNode* cur = nodes.front();
            if (cur->right) nodes.push(cur->right);
            if (cur->left) nodes.push(cur->left);  
            nodes.pop();
        }
    }
    return res;
    }
};

前序遍历，比较简洁

class Solution {
public:
    void dfs(TreeNode* root, int lv, vector<int> &res){
        if(!root)   return;
        if(lv>=res.size()) res.push_back(root->val);
        dfs(root->right,lv+1,res);
        dfs(root->left,lv+1,res);
    }

    vector<int> rightSideView(TreeNode* root) {
        vector<int> res;
        dfs(root, 0, res);
        return res;
    }
};

236. Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
求任意二叉树的最小公共结点

思路并不复杂，思路就是记录从根节点达到p和q的两条路径，显然知道了两条路径之后，不相同的上一个节点就是最低公共祖先。

/**
 * 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<TreeNode*> qPath;  
    vector<TreeNode*> pPath; 
    void findPath(TreeNode* node, TreeNode* p, TreeNode* q, vector<TreeNode*> & res){
        if (!pPath.empty() && !qPath.empty()) return;
        res.push_back(node);
        if (node == p) pPath = res;
        if (node == q) qPath = res;
        if (node -> left) findPath(node->left, p, q, res);
        if (node -> right) findPath(node->right, p, q, res);
        res.pop_back(); // back_track
    }
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(!root || !p || !q) return NULL;  
        vector<TreeNode*> res;//找出两个包含这两个结点的路径  
        findPath(root,q,p,res); 
        if (pPath.empty() || qPath.empty()) return NULL;// 有一个结点没有查找到
        int i = 0, j = 0;
        int pLen = pPath.size(), qLen = qPath.size();
        while (i < pLen && j < qLen) {
            if (pPath[i] != qPath[j]) return pPath[i-1];
            if (i == pLen - 1) return pPath[i];
            if (j == qLen - 1) return qPath[j];
            ++i, ++j;
        }
    }
};

别人的DFS的简洁的写法，但是我质疑的一点就是，如果两个节点有一个不在这个树上的时候，这样怎么处理？

TreeNode * dfsTraverse(TreeNode * root, TreeNode * p , TreeNode * q)
{
    if( root == p || root == q || root == NULL)
        return root;
    TreeNode * parent1 = dfsTraverse(root->left, p, q);
    TreeNode * parent2 = dfsTraverse(root->right, p, q);
    if( parent1 && parent2)
        return root;
    else
        return parent1 ? parent1:parent2;
}
TreeNode * lowestCommonAncestor(TreeNode * root, TreeNode * p, TreeNode * q)
{
    return dfsTraverse(root, p, q);
}