Binary Tree
Definition： at most two children node.
Binary Tree Example:
10 ==root
/ \
13 15 cur
/ \ / \
21 72 12 2
/ \
null null
class TreeNode{
int value;
TreeNode * left;
TreeNode * right;
TreeNode * parent //point to this node's parent node.
}
面试常见题型：
基本知识点1: tree traverse
1. pre-order.
2.in-order.
3.post-order.
关键点：base case通常就是叶子节点下面的空节点。
Balanced binary tree:
对于树的任意一个节点，左子树和右子树的高度差不超过1。
Complete binary tree(完全二叉树)
底层是一个数组，数组内部的数字必须是连续的，不能有空余的内存空间。
Binary Search Tree（二叉查找树）
10
/ \
5 15
/ \ / \
2 7 12 20
注意：对于根节点10，必须整个左子树（左子树上的所有节点）都必须比10小，整个右子树（右子树上的所有节点）必须比10大。
同时binary search tree不允许有重复的node;
Binary tree 往往是最常见的和recursion结合最紧密的面试题目类型。
理由：
1.每层的node所具备的性质，传递的值和下一层的性质往往一致，比较容易定义recursive rule。
2.base case: null pointer under the leaf node.
3.Example1:int getHeight(Node root)
4.Example2:统计tree里面有多少个node。
常见面试题型：
How to get integer value(height) for a problem with size = n? But how?
GetHeight of a binary tree?
public int getHeight(TreeNode root){
if(root == null){
return 0;
}
int left = getHeight(root.left);
int right = getHeight(root.right);
return 1 + Math.max(left,right);
}
Time = O(n);
space = O(n) == O(height);
---------------------------------------------------------------------------------------------------------------------------
Q3:How to determine whether a binary tree is a balanced binary tree?
public boolean isBalanced(TreeNode tree){
if(root == null){
return true;
}
int left = getHeight(root.left);
int right = getHeight(root.right);
if(Math.abs(left - right) > 1){
return false;
}
return isBalanced(root.left) && isBalanced(root.right);
}
时间复杂度分析：
isBalanced(n/2 + n/2)
/ \
getHeight getHeight
(n/4 + n/4) (n/4 + n/4)
因为是一个平衡二叉树所以：层数logn So: Time : O(nlogn)
---------------------------------------------------------------------------------------------------------------------------
Q4:怎么判断一颗二叉树左右两边是不是对称的？
10
5a | 5b
1a 3a | 3b 1b
2a4a 6a8a | 8b6b 4b2b
public boolean isSymmetric(TreeNode noe,TreeNode two){
if(one == null && two == null){
return true;
}
if(one ==null || two == null){
return false;
}
if(one.value == two.value){
return false;
}
return isSymmetric(one.left,two.right) && isSymmetric(one.right,one.left);
}
Time = O(n);
space = O(n) -> O(height) if the tree is balaced -> O(logn)
---------------------------------------------------------------------------------------------------------------------------
Q5: