树常见的算法操作

树是数据结构中很重要的一部分，也是各大公司面试常考部分。

继树的各种遍历算法之后，今天又整理一下树的常见算法操作。

本文包括:

1.求节点的最近公共祖先

2.树的序列化与反序列化

3.已知先序遍历和中序遍历构造二叉树

4.已知中序遍历和后序遍历构造二叉树

1.求节点最近的公共祖先

此题不同的要求有不同的解法

如果已知树中的每一个结点有指向父节点的指针：

思路:从给定节点遍历到根节点，当父节点相等时返回。

解法1

private ArrayList<TreeNode> getPath2Root(TreeNode node) {
ArrayList<TreeNode> list = new ArrayList<TreeNode>();
while (node != null) {
list.add(node);
node = node.parent;
}
return list;
}
public TreeNode lowestCommonAncestor(TreeNode node1, TreeNode node2) {
ArrayList<TreeNode> list1 = getPath2Root(node1);
ArrayList<TreeNode> list2 = getPath2Root(node2);

int i, j;
for (i = list1.size() - 1, j = list2.size() - 1; i >= 0 && j >= 0; i--, j--) {
if (list1.get(i) != list2.get(j)) {
return list1.get(i).parent;
}
}
return list1.get(i+1);
}

解法2:

private TreeNode getRoot(node) {
while (node.parent != null) {
node = node.parent;
}
return node;
}

private TreeNode getAncestor(TreeNode root, TreeNode node1, TreeNode node2) {
if (root == null || root == node1 || root == node2) {
return root;
}

TreeNode left = getAncestor(root.left, node1, node2);
TreeNode right = getAncestor(root.right, node1, node2);

if (left != null && right != null) {
return root;
}
if (left != null) {
return left;
}
if (right != null) {
return right;
}
return null;
}

public TreeNode lowestCommonAncestor(TreeNode node1, TreeNode node2) {
if (node1 == null || node2 == null) {
return null;
}
TreeNode root = getRoot(node1);
return getAncestor(root, node1, node2);
}

如果树中的节点不带有指向父节点的指针:

思路利用先序遍历和中序遍历的特点，利用先序遍历分辨根节点，根据根节点和给定节点AB的位置关系来确定公共祖先，根据中序遍历结果，两个节点中间的节点即为所求祖先，若中间没有节点，则中序遍历中下标较小的即为祖先。

TreeNode node =null;
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode A, TreeNode B) {
// write your code here

TreeNode result =null;
ArrayList<TreeNode> inorder = new ArrayList<TreeNode>();
ArrayList<TreeNode> preorder = new ArrayList<TreeNode>();
inorderto(root,inorder);
preorderto(root,preorder);
//得到AB节点在中序遍历中的下标
int aValue = inorder.indexOf(A);
int bValue = inorder.indexOf(B);
int lastindex = 0 ;
for(int i=0;i<preorder.size();){
int rootValue = inorder.indexOf(preorder.get(i));
if((aValue<=rootValue && bValue>=rootValue)||(aValue>=rootValue && bValue<=rootValue)){
result = preorder.get(i);
break;
}

else
i++;
}
return result;
}
public void inorderto(TreeNode root,ArrayList<TreeNode> list){
if(root==null)
return;
else{
inorderto(root.left,list);
list.add(root);
inorderto(root.right,list);
}
}
public void preorderto(TreeNode root,ArrayList<TreeNode> list){
if(root==null)
return;
else{
list.add(root);
preorderto(root.left,list);
preorderto(root.right,list);
}
}

2.树的序列化与反序列化

将树的节点存入文件，再从文件读出构造出树称为树的序列化与反序列化。

思路:利用先序遍历(其他遍历亦可)，将树的节点值存入字符串，若节点为空则存入#，每个节点之间用','隔开用来区分数字。

读出时亦用先序遍历的方法。

public String serialize(TreeNode root) {
// write your code here
String str = "";
Stack<TreeNode> stack = new Stack<TreeNode>();
if(root!=null)
stack.push(root);
while(!stack.isEmpty()){
TreeNode node = stack.pop();
if(node!=null){
str += node.val + ",";
stack.push(node.right);
stack.push(node.left);
}
else
str += "#,";
}
return str;
}
int index =0 ;
public TreeNode deserialize(String data) {
if(data=="")
return null;
String val = outValue(index,data);
index += 1;
if(val.equals("#"))
return null;
else{
TreeNode node = new TreeNode(Integer.parseInt(val));

node.left = deserialize(data);
node.right = deserialize(data);
return node;
}
}

public String outValue(int index,String str){
String res = "";
for(int i=0,j=0;i<str.length();i++){
if(str.charAt(i)==',')
j++;
else if(j==index)
res += str.charAt(i);
if(j>index)
break;
}
return res;
}

3.已知先序遍历和中序遍历构造二叉树

思路:利用先序遍历区分根节点，利用中序遍历区分左右子树

private int findPosition(int[] arr, int start, int end, int key) {
int i;
for (i = start; i <= end; i++) {
if (arr[i] == key) {
return i;
}
}
return -1;
}

private TreeNode myBuildTree(int[] inorder, int instart, int inend,
int[] preorder, int prestart, int preend) {
if (instart > inend) {
return null;
}

TreeNode root = new TreeNode(preorder[prestart]);
int position = findPosition(inorder, instart, inend, preorder[prestart]);

root.left = myBuildTree(inorder, instart, position - 1,
preorder, prestart + 1, prestart + position - instart);
root.right = myBuildTree(inorder, position + 1, inend,
preorder, position - inend + preend + 1, preend);
return root;
}

public TreeNode buildTree(int[] preorder, int[] inorder) {
if (inorder.length != preorder.length) {
return null;
}
return myBuildTree(inorder, 0, inorder.length - 1, preorder, 0, preorder.length - 1);
}

4.已知中序遍历和后序遍历构造二叉树

思路：和前者思路差不多，利用后序遍历来寻找根节点，利用中序遍历分辨左右子树

public TreeNode buildTree(int[] inorder, int[] postorder) {
// write your code here
if(inorder.length != postorder.length)
return null;
return myTree(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
}
public TreeNode myTree(int[]inorder,int instart,int inend,
int[] postorder,int poststart,int postend){
if(instart>inend)
return null;
TreeNode root = new TreeNode(postorder[postend]);
int position = findPosition(inorder,instart,inend,postorder[postend]);
root.left = myTree(inorder,instart,position-1,postorder,poststart,poststart+position-1-instart);
root.right = myTree(inorder,position+1,inend,postorder,position - inend + postend ,postend-1);
return root;
}
private int findPosition(int[] arr, int start, int end, int key) {
int i;
for (i = start; i <= end; i++) {
if (arr[i] == key) {
return i;
}
}
return -1;
}