leetcode Ch4-Binary Tree & BFS & Divide/Conquer

一、 

1. Lowest Common Ancestor

class Solution {
public:
    TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *A, TreeNode *B) {
        if (root == NULL || root == A || root == B) {
            return root;
        }
        TreeNode* left = lowestCommonAncestor(root->left, A, B);
        TreeNode* right = lowestCommonAncestor(root->right, A, B);
        if (left != NULL && right != NULL) {
            return root;
        }
        if (left != NULL) {
            return left;
        }
        if (right != NULL) {
            return right;
        }
        return NULL;
    }
};

refer : July，剑指offer

2. Lowest Common Ancestor of a Binary Search Tree

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

二. Level order [BFS]

1. Binary Tree Level Order Traversal

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if (root == NULL) {
            return result;
        }
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            int size = q.size();
            vector<int> v;
            for (int i = 0; i < size; i++) {
                TreeNode* tmp = q.front();
                q.pop();
                v.push_back(tmp->val);
                if (tmp->left != NULL) { 
                    q.push(tmp->left);
                }
                if (tmp->right != NULL) {
                    q.push(tmp->right);
                }
            }
            result.push_back(v);
        }
        return result;
    }
};

2. Binary Tree Level Order Traversal II

class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        if (root == NULL) {
            return result;
        }
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            int size = q.size();
            vector<int> v;
            for (int i = 0; i < size; i++) {
                TreeNode* tmp = q.front();
                q.pop();
                v.push_back(tmp->val);
                if (tmp->left != NULL) { 
                    q.push(tmp->left);
                }
                if (tmp->right != NULL) {
                    q.push(tmp->right);
                }
            }
            result.push_back(v);
        }
        reverse(result.begin(), result.end());
        return result;
    }
};

在1的基础上多加一句reverse即可。

3. Binary Tree Zigzag Level Order Traversal

class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if (root == NULL) {
            return result;
        }
        queue<TreeNode*> q;
        q.push(root);
        int count = 0;
        while(!q.empty()) {
            count++;
            int size = q.size();
            vector<int> v;
            for (int i = 0; i < size; i++) {
                TreeNode* tmp = q.front();
                q.pop();
                v.push_back(tmp->val);
                if (tmp->left != NULL) { 
                    q.push(tmp->left);
                }
                if (tmp->right != NULL) {
                    q.push(tmp->right);
                }
            }
            if (count % 2 == 0) {
                reverse(v.begin(), v.end());    
            }
            result.push_back(v);
        }
        return result;
    }
};

在1的基础上多加个count变量，偶数行就reverse一下即可

三、

1. Insert Node in a Binary Search Tree

TreeNode* insertNode(TreeNode* root, TreeNode* node) {
    if (root == NULL) {
        return node;
    }
    if (node->val > root->val) {
        root->right = insertNode(root->right, node);
    } else {
        root->left = insertNode(root->left, node);
    }
    return root;
}

2. Search Range in Binary Search Tree

code1: 

class Solution {
public:
    vector<int> searchRange(TreeNode* root, int k1, int k2) {
        helper(root, k1, k2);
        return result;
    }
    void helper(TreeNode* root, int k1, int k2) {
        if (root == NULL) {
            return;
        }
        if (k1 < root->val) {//说明左子树里有可能有
            helper(root->left, k1, k2);
        }
        if (root->val >= k1 && root->val <= k2) {
            result.push_back(root->val);
        }
        if (k2 > root->val) {
            helper(root->right, k1, k2);
        }
    }
private:
    vector<int>    result;
};

code2: 自己实现的，太繁琐。

vector<int> searchRange(TreeNode* root, int k1, int k2) {
    vector<int> result;
    if (root == NULL) {
        return result;
    }
    if (root->val < k1) {
        return searchRange(root->right, k1, k2);
    }
    if (root->val > k2) {
        return searchRange(root->left, k1, k2);
    }
    if (root->val >= k1 && root->val <= k2) {
        vector<int> tmp1 = searchRange(root->left, k1, root->val - 1);
        vector<int> tmp2 = searchRange(root->right, root->val + 1, k2);
        result.insert(result.end(), tmp1.begin(), tmp1.end());
        result.push_back(root->val);
        result.insert(result.end(), tmp2.begin(), tmp2.end());
    }
    return result;
}

Binary Search Tree Iterator

class BSTIterator {
public:
    BSTIterator(TreeNode* root) {
        pushAll(root);
    }
    
    bool hasNext() {
        return (!myStack.empty());
    }

    int next() {
        TreeNode* tmp = myStack.top();
        myStack.pop();
        pushAll(tmp->right);
        return tmp->val;        
    }
    
private:
    stack<TreeNode*> myStack;
    void pushAll(TreeNode* node);
};

void BSTIterator::pushAll(TreeNode* node) {
    while (node != NULL) {
        myStack.push(node);
        node = node->left;
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */

Remove Node in Binary Search Tree

Binary Tree Maximum Path Sum 

参见 ref 十五

Binary Tree Serialization

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对于n个数的数组，一个数x如果从左往右数是第k个数，那么从右往左数的话是第(n - k + 1)个数。