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。

  1 struct TreeNode {
  2     TreeNode *left;
  3     TreeNode *right;
  4     TreeNode *parent;
  5     int val;
  6     TreeNode(int v):val(v),left(NULL),right(NULL),parent(NULL) {}
  7 };
  8
  9 // 4.6
 10 TreeNode* findInorderNext(TreeNode* node) {
 11     if (node == NULL) {
 12         return NULL;
 13     } else if (node->right) {
 14         node = node->right;
 15         while (node->left) node = node->left;
 16         return node;
 17     } else {
 18         while (node->parent && node->parent->left != node) node = node->parent;
 19         return node->parent;
 20     }
 21 }
 22
 23 TreeNode* findPreordeNext(TreeNode* node) {
 24     if (node == NULL) {
 25         return NULL;
 26     } else if (node->left) {
 27         return node->left;
 28     } else if (node->right) {
 29         return node->right;
 30     } else {
 31         while (node->parent && node->parent->right == node) {
 32             node = node->parent;
 33         }
 34         if (node->parent) return node->parent->right;
 35         else return NULL;
 36     }
 37 }
 38
 39 TreeNode* findPostorderNext(TreeNode* node) {
 40     if (node == NULL || node->parent == NULL) {
 41         return NULL;
 42     } else if (node->parent->right == node || node->parent->right == NULL) { // right subtree
 43         return node->parent;
 44     } else {
 45         node = node->parent->right;
 46         while (node->left) node = node->left;
 47         return node;
 48     }
 49 }
 50
 51
 52 // 4.7 (1) p, q in the BST, O(h) runtime
 53 TreeNode* LCAInBST(TreeNode *root, TreeNode *p, TreeNode *q) {
 54     if (!root || !p || !q) return NULL;
 55     if (max(p->val, q->val) < root->val) {
 56         return LCAInBST(root->left, p, q);
 57     } else if (min(p->val, q->val) > root->val) {
 58         return LCAInBST(root->right, p, q);
 59     } else {
 60         return root;
 61     }
 62 }
 63
 64 // 4.7 (2.1) p, q may not in the tree, this is just a binary tree
 65 int matchCount(bool pExisted, bool qExisted) {
 66     int m = 0;
 67     if (pExisted) m++;
 68     if (qExisted) m++;
 69     return m;
 70 }
 71 TreeNode* LCAInBT(TreeNode *root, TreeNode *p, TreeNode *q, bool &pExisted, bool &qExisted) {
 72     if (!root) return NULL;
 73
 74     TreeNode* ret = LCAInBT(root->left, p, q, pExisted, qExisted);
 75     int lm = matchCount(pExisted, qExisted);
 76     if (lm == 2) return ret; // p, q are in the left subtree
 77     ret = LCAInBT(root->right, p, q, pExisted, qExisted);
 78     int rm = matchCount(pExisted, qExisted);
 79     if (rm == 2) { // p, q are found
 80         if (lm == 1) return root; //p, q are in different subtree, thus return root
 81         else return ret; // lm == 0, p, q are in the right subtree
 82     }
 83     if (root == p) pExisted = true;
 84     if (root == q) qExisted = true;
 85     if (pExisted && qExisted) return root; // if q is a child of q (or, q is a child of p)
 86     return NULL; // if p or q is not existed
 87 }
 88
 89 // 4.7(2.2) the parent node is available
 90 int getHeight(TreeNode* p) {
 91     int h = 0;
 92     while (p) {
 93         p = p->parent;
 94         h++;
 95     }
 96     return h;
 97 }
 98
 99 TreeNode* LCAInBT(TreeNode *p, TreeNode *q) {
100     if (!p || !q) return NULL;
101     int h1 = getHeight(p);
102     int h2 = getHeight(q);
103     if (h1 > h2) {
104         swap(p, q);
105         swap(h1, h2);
106     }
107     for (int i = 0; i < h2 - h1; ++i) {
108         q = q->parent;
109     }
110
111     while (p && q) {
112         if (p == q) return p;
113         p = p->parent;
114         q = q->parent;
115     }
116     return NULL;
117 }

转载于:https://www.cnblogs.com/linyx/p/3782607.html

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