Careercup | Chapter 4

二叉查换树，左孩子小于等于根，右孩子大于根。 

完全二叉树，除最后一层外，每一层上的节点数均达到最大值；在最后一层上只缺少右边的若干结点。 complete binary tree

满二叉树，完美二叉树是全部结点数达到最大的二叉树。perfect binary tree, full binary tree.

平衡二叉树，左右子树的高度在一定范围内。 

二叉查找树（Binary Search Tree），也称二叉搜索树、有序二叉树（ordered binary tree），排序二叉树（sorted binary tree），是指一棵空树或者具有下列性质的二叉树：

1. 若任意节点的左子树不空，则左子树上所有结点的值均小于或等于它的根结点的值；
2. 任意节点的右子树不空，则右子树上所有结点的值均大于它的根结点的值；
3. 任意节点的左、右子树也分别为二叉查找树。
4. 没有键值相等的节点（no duplicate nodes）。

在二叉查找树删去一个结点，分三种情况讨论：

• 若*p结点为叶子结点，即PL(左子树)和PR(右子树)均为空树。由于删去叶子结点不破坏整棵树的结构，则只需修改其双亲结点的指针即可。
• 若*p结点只有左子树PL或右子树PR，此时只要令PL或PR直接成为其双亲结点*f的左子树（当*p是左子树）或右子树（当*p是右子树）即可，作此修改也不破坏二叉查找树的特性。
• 若*p结点的左子树和右子树均不空。在删去*p之后，为保持其它元素之间的相对位置不变，可按中序遍历保持有序进行调整，可以有两种做法：其一是令*p的左子树为*f的左/右(依*p是*f的左子树还是右子树而定)子树，*s为*p左子树的最右下的结点，而*p的右子树为*s的右子树；其二是令*p的直接前驱（in-order predecessor）或直接后继（in-order successor）替代*p，然后再从二叉查找树中删去它的直接前驱（或直接后继）。

4.1 Implement a function to check if a binary tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that the heights of the two subtrees of any node never differ by more than one.

点此。

4.2 Given a directed graph, design an algorithm to find out whether there is a route between two nodes.

BFS或者DFS就行了，不过BFS可以避免在单条路径上挖得太深。

4.3 Given a sorted (increasing order) array with unique integer elements, write an algorithm to create a binary search tree with minimal height

点此。

4.4 Given a binary tree, design an algorithm which creates a linked list of all the nodes at each depth (e.g., if you have a tree with depth D, you'll have D linked lists).

BFS或者DFS都可以。注意的是，DFS虽然会用额外的O(lgn)的栈空间，但是整体的空间复杂度还是O(n)。

4.5 Implement a function to check if a binary tree is a binary search tree.

点此。BST要和中序遍历联系在一起。

4.6  Write an algorithm to find the 'next'node (i.e., in-order successor) of a given node in a binary search tree. You may assume that each node has a link to its parent.

如果有root结点，直接用中序遍历记录pre结点就可以做的。如果只有结点，那么需要分情况。

4.7 Design an algorithm and write code to find the first common ancestor of two nodes in a binary tree. Avoid storing additional nodes in a data structure. NOTE: This is not necessarily a binary search tree.

leetcode上的博文，讲得很仔细。这道题很经典，尤其是有parent结点时的解法，类似求intersect linklist的交叉点。

4.8 You have two very large binary trees: Tl, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of Tl. A tree T2 is a subtree of Tl if there exists a node n in Tl such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.

这道题可以穷搜，也可以基于pre-order和in-order的travesal串，用子串查找。

4.9 You are given a binary tree in which each node contains a value. Design an algorithm to print all paths which sum to a given value. The path does not need to start or end at the root or a leaf.

只能穷搜了。

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

// 4.6
TreeNode* findInorderNext(TreeNode* node) {
    if (node == NULL) {
        return NULL;
    } else if (node->right) {
        node = node->right;
        while (node->left) node = node->left;
        return node;
    } else {
        while (node->parent && node->parent->left != node) node = node->parent;
        return node->parent;
    }
}

TreeNode* findPreordeNext(TreeNode* node) {
    if (node == NULL) {
        return NULL;
    } else if (node->left) {
        return node->left;
    } else if (node->right) {
        return node->right;
    } else {
        while (node->parent && node->parent->right == node) {
            node = node->parent;
        }
        if (node->parent) return node->parent->right;
        else return NULL;
    }
}

TreeNode* findPostorderNext(TreeNode* node) {
    if (node == NULL || node->parent == NULL) {
        return NULL;
    } else if (node->parent->right == node || node->parent->right == NULL) { // right subtree
        return node->parent;
    } else {
        node = node->parent->right;
        while (node->left) node = node->left;
        return node;
    }
}


// 4.7 (1) p, q in the BST, O(h) runtime
TreeNode* LCAInBST(TreeNode *root, TreeNode *p, TreeNode *q) {
    if (!root || !p || !q) return NULL;
    if (max(p->val, q->val) < root->val) {
        return LCAInBST(root->left, p, q);
    } else if (min(p->val, q->val) > root->val) {
        return LCAInBST(root->right, p, q);
    } else {
        return root;
    }
}

// 4.7 (2.1) p, q may not in the tree, this is just a binary tree
int matchCount(bool pExisted, bool qExisted) {
    int m = 0;
    if (pExisted) m++;
    if (qExisted) m++;
    return m;
}
TreeNode* LCAInBT(TreeNode *root, TreeNode *p, TreeNode *q, bool &pExisted, bool &qExisted) {
    if (!root) return NULL;

    TreeNode* ret = LCAInBT(root->left, p, q, pExisted, qExisted);
    int lm = matchCount(pExisted, qExisted);
    if (lm == 2) return ret; // p, q are in the left subtree
    ret = LCAInBT(root->right, p, q, pExisted, qExisted);
    int rm = matchCount(pExisted, qExisted);
    if (rm == 2) { // p, q are found
        if (lm == 1) return root; //p, q are in different subtree, thus return root
        else return ret; // lm == 0, p, q are in the right subtree
    }
    if (root == p) pExisted = true;
    if (root == q) qExisted = true;
    if (pExisted && qExisted) return root; // if q is a child of q (or, q is a child of p)
    return NULL; // if p or q is not existed 
}

// 4.7(2.2) the parent node is available
int getHeight(TreeNode* p) {
    int h = 0;
    while (p) {
        p = p->parent;
        h++;
    }
    return h;
}

TreeNode* LCAInBT(TreeNode *p, TreeNode *q) {
    if (!p || !q) return NULL;
    int h1 = getHeight(p);
    int h2 = getHeight(q);
    if (h1 > h2) {
        swap(p, q);
        swap(h1, h2);
    }
    for (int i = 0; i < h2 - h1; ++i) {
        q = q->parent;
    }

    while (p && q) {
        if (p == q) return p;
        p = p->parent;
        q = q->parent;
    }
    return NULL;
}