Leetcode 二叉树 106 105 654 617 700 98 530

106. Construct Binary Tree from Inorder and Postorder Traversal

class Solution {
public:
    TreeNode* buildTree(vector& inorder, vector& postorder) {
        if(postorder.size() == 0) return NULL;
        int rootVal = postorder[postorder.size()-1];
        TreeNode* root = new TreeNode(rootVal);

        if(postorder.size() == 1) return root;

        int delimiterIndex;
        for(delimiterIndex = 0; delimiterIndex < inorder.size(); delimiterIndex++){
            if(inorder[delimiterIndex] == rootVal){
                break;
            }
        }

        vector leftinorder(inorder.begin(), inorder.begin()+delimiterIndex);
        vector rightinorder(inorder.begin()+delimiterIndex+1, inorder.end());
        postorder.resize(postorder.size()-1);
        vector leftpostorder(postorder.begin(), postorder.begin()+leftinorder.size());
        vector rightpostorder(postorder.begin()+leftinorder.size(), postorder.end());

        root->left = buildTree(leftinorder, leftpostorder);
        root->right = buildTree(rightinorder, rightpostorder);
        return root;
    }
};

直接用下标写

class Solution {
public:
    TreeNode* buildTree(vector& inorder, vector& postorder) {
        if(inorder.size() == 0) return NULL;
        return traversal(inorder, 0, inorder.size(), postorder, 0, postorder.size()); 

    }
private:
    TreeNode* traversal(vector& inorder, int inorderBegin, int inorderEnd, vector& postorder, int postorderBegin, int postorderEnd){
        if(postorderBegin == postorderEnd) return NULL;

        int rootVal = postorder[postorderEnd-1];
        TreeNode* root = new TreeNode(rootVal);

        if(postorderEnd - postorderBegin == 1) return root;

        int delimiterIndex;
        for(delimiterIndex = inorderBegin; delimiterIndex < inorderEnd; delimiterIndex++){
            if(inorder[delimiterIndex] == rootVal){
                break;
            }
        }

        int leftInorderBegin = inorderBegin;
        int leftInorderEnd = delimiterIndex;
        int rightInorderBegin = delimiterIndex + 1;
        int rightInorderEnd = inorderEnd;

        int leftPostorderBegin = postorderBegin;
        int leftPostorderEnd = postorderBegin + delimiterIndex - inorderBegin;
        int rightPostorderBegin = postorderBegin + delimiterIndex - inorderBegin;
        int rightPostorderEnd = postorderEnd - 1;

        root->left = traversal(inorder, leftInorderBegin, leftInorderEnd, postorder, leftPostorderBegin, leftPostorderEnd);
        root->right = traversal(inorder, rightInorderBegin, rightInorderEnd, postorder, rightPostorderBegin, rightPostorderEnd);
        return root;
    }
};

105. Construct Binary Tree from Preorder and Inorder Traversal

class Solution {
public:
    TreeNode* buildTree(vector& preorder, vector& inorder) {
        if(preorder.size() == 0) return NULL;
        return traversal(preorder, 0, preorder.size(), inorder, 0, inorder.size());

    }

    TreeNode* traversal(vector& preorder, int preorderBegin, int preorderEnd, vector inorder, int inorderBegin, int inorderEnd){
        if(preorderBegin - preorderEnd == 0) return NULL;

        int rootVal = preorder[preorderBegin];
        TreeNode* root = new TreeNode(rootVal);

        if(preorderEnd - preorderBegin == 1) return root;

        int delimiterIndex;
        for(delimiterIndex = inorderBegin; delimiterIndex < inorderEnd; delimiterIndex++){
            if(inorder[delimiterIndex] == rootVal) break;
        }

        int leftInorderBegin = inorderBegin;
        int leftInorderEnd = delimiterIndex;
        int rightInderBegin = delimiterIndex+1;
        int rightInderEnd = inorderEnd;

        int leftPreorderBegin = preorderBegin + 1;
        int leftPreorderEnd = leftPreorderBegin + leftInorderEnd - leftInorderBegin;
        int rightPreorderBegin = leftPreorderEnd;
        int rightPreorderEnd = preorderEnd;

        root->left = traversal(preorder, leftPreorderBegin, leftPreorderEnd, inorder, leftInorderBegin, leftInorderEnd);
        root->right = traversal(preorder, rightPreorderBegin, rightPreorderEnd, inorder, rightInderBegin, rightInderEnd);
        return root;

    }
}

 654. Maximum Binary Tree

class Solution {
public:
    TreeNode* constructMaximumBinaryTree(vector& nums) {
        return traversal(nums, 0, nums.size());

    }

    TreeNode* traversal(vector& nums, int begin, int end){
        if(end-begin == 0) return NULL;
        int maxVal = INT_MIN;
        int maxIndex; 
        for(int i=begin; ileft = traversal(nums, leftBegin, leftEnd);
        root->right = traversal(nums, rightBegin, rightEnd);
        return root;
    }
};

617. Merge Two Binary Trees

class Solution {
public:
    TreeNode* mergeTrees(TreeNode* root1, TreeNode* root2) {
        if(!root1) return root2;
        if(!root2) return root1;

        root1->val += root2->val;
        root1->left = mergeTrees(root1->left, root2->left);
        root1->right = mergeTrees(root1->right, root2->right);
        return root1;
    }
};

 700. Search in a Binary Search Tree

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if(root == NULL) return NULL;
        if(root->val == val) return root;
        if(root->val > val) return searchBST(root->left, val);
        if(root->val < val) return searchBST(root->right, val);
        return NULL;
    }
};

要有root == NULL的终止条件

最后要有return NULL，因为没有写else， 所以默认会有其他情况

如果不写return NULL也可以这么写

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        if(root == NULL) return NULL;
        if(root->val == val) return root;
        if(root->val > val) return searchBST(root->left, val);
        else return searchBST(root->right, val);
        
    }
};

2.迭代法

class Solution {
public:
    TreeNode* searchBST(TreeNode* root, int val) {
        while(root != NULL){
            if(root->val == val) return root;
            else if(root->val > val) root = root->left;
            else root = root->right;
        }
        return NULL;
    }
};

98. Validate Binary Search Tree

两个方法思想一样，一个先提取出生成一个数组， 一个直接递归

class Solution {
private:
    vector res;
    void traversal(TreeNode* root){
        if(root == NULL) return;
        traversal(root->left);
        res.push_back(root->val);
        traversal(root->right);
    }
public:
    bool isValidBST(TreeNode* root) {
        res.clear();
        traversal(root);
        for(int i=1; i

class Solution {
public:
    TreeNode* pre = NULL;
    bool isValidBST(TreeNode* root) {
        if(root == NULL) return true;
        bool left = isValidBST(root->left);
        if(pre != NULL && pre->val >= root->val) return false;
        pre = root;
        bool right = isValidBST(root->right);
        return left && right;
    }
};

530. Minimum Absolute Difference in BST

跟前一道题的两种想法比较像

class Solution {
private:
    vector res;
    void traversal(TreeNode* root){
        if(root == NULL) return;
        traversal(root->left);
        res.push_back(root->val);
        traversal(root->right);
    }
public:
    int getMinimumDifference(TreeNode* root) {
        res.clear();
        traversal(root);
        int minDif = INT_MAX;
        for(int i=1; i

class Solution {
private:
    TreeNode* pre = NULL;
    int minDif = INT_MAX;
    void traversal(TreeNode* root){
        if(root == NULL) return;
        traversal(root->left);
        if(pre != NULL){
            int dif = root->val - pre->val;
            if(dif < minDif) minDif = dif;
        }
        pre = root;
        traversal(root->right);
    }

public:
    int getMinimumDifference(TreeNode* root) {
        traversal(root);
        return minDif;
    }
};