Data Structure Notes--Binary Tree Classic High Frequency Questions

1--The nearest common ancestor of the binary tree

Main idea:

        There are only two cases of the nearest ancestor: ① bottom-up, when the two destination nodes are in the left and right subtrees of the current node, the current node is the nearest ancestor of the two destination nodes; ② the nearest ancestor and one of the destination nodes points are the same, another destination node is on the subtree of the destination node;

        Recursively search for the destination node, when the destination node is found, return to the destination node, otherwise return NULL; when a node finds two destination nodes on the left and right subtrees, it indicates that this node is the nearest ancestor; otherwise Return the return node of the subtree that is not empty (then the two nodes correspond to the second case);

#include <iostream>
#include <vector>
#include <stack>

struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == NULL || root->val == p->val || root->val == q->val) return root;

        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        if(left != NULL && right != NULL) return root;
        else if( left != NULL) return left;
        else return right;
    }
};

int main(int argc, char *argv[]){

    TreeNode *Node1 = new TreeNode(3);
    TreeNode *Node2 = new TreeNode(5);
    TreeNode *Node3 = new TreeNode(1);
    TreeNode *Node4 = new TreeNode(6);
    TreeNode *Node5 = new TreeNode(2);
    TreeNode *Node6 = new TreeNode(0);
    TreeNode *Node7 = new TreeNode(8);
    TreeNode *Node8 = new TreeNode(7);
    TreeNode *Node9 = new TreeNode(4);

    Node1->left = Node2;
    Node1->right = Node3;
    Node2->left = Node4;
    Node2->right = Node5;
    Node3->left = Node6;
    Node3->right = Node7;
    Node5->left = Node8;
    Node6->right = Node9;

    Solution S1;
    TreeNode* res = S1.lowestCommonAncestor(Node1, Node2, Node9);
    std::cout << res->val << std::endl;

    return 0;
}

2--Inorder successor node of binary search tree

Main idea:

        If the p node has a right subtree, return the leftmost node of its right subtree (definition of inorder traversal);

        If the p node has no right subtree, start looking for the parent node of p node from the root node; (according to the definition of binary search tree, you can save the time of searching, just search on one side);

#include <iostream>
#include <vector>
#include <stack>


struct TreeNode {
     int val;
     TreeNode *left;
     TreeNode *right;
     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        TreeNode *res = NULL;
        if(p->right != NULL){ // p的中序后继是其右子树上最左的结点（即右字数上最先返回的结点）
            res = p->right;
            while(res->left != NULL) res = res->left;
            return res;
        }

        // p没有右子树，从root结点开始搜索p的父亲结点
        while(root != NULL){
            if(root->val > p->val){ // p在左子树上
                res = root;
                root = root->left; // 在左子树上找到最后一个比p大的结点（中序遍历是有序的，中序后继结点表明是比p结点大）
            }
            else{
                root = root->right; // p在右子树上
            }
        }
        return res;
    }
};

int main(int argc, char *argv[]){

    TreeNode *Node1 = new TreeNode(2);
    TreeNode *Node2 = new TreeNode(1);
    TreeNode *Node3 = new TreeNode(3);

    Node1->left = Node2;
    Node1->right = Node3;

    Solution S1;
    TreeNode* res = S1.inorderSuccessor(Node1, Node2);
    std::cout << res->val << std::endl;

    return 0;
}

3--Serialization and deserialization of binary tree

Main idea: