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：
- 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；
- 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；
- 任意节点的左、右子树也分别为二叉查找树。
- 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。
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。
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。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 }