二叉树相关
有一棵二叉树，请设计一个算法，按照层次打印这棵二叉树。
给定二叉树的根结点root，请返回打印结果，结果按照每一层一个数组进行储存，所有数组的顺序按照层数从上往下，且每一层的数组内元素按照从左往右排列。保证结点数小于等于500。
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val){
this.val = val;
}
}
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
public class TreePrinter {
public static int[][] printTree(TreeNode root){
TreeNode last; //指向当前节点
TreeNode nlast; //指向当前节点下一行的最后一个节点
last = root;
nlast = last;
Queue<TreeNode> que = new LinkedList<TreeNode>();
que.offer(root);
ArrayList<ArrayList<Integer>> store = new ArrayList<ArrayList<Integer>>();
store.add(new ArrayList<Integer>());
int hight = 0;
while(!que.isEmpty()){
TreeNode out = que.poll();
//System.out.print(out.val+" ");
store.get(hight).add(out.val);
if(out.left!=null){
que.offer(out.left);
nlast = out.left;
}
if(out.right!=null){
que.offer(out.right);
nlast = out.right;
}
if(out==last){
//System.out.println();
last = nlast;
hight++;
store.add(new ArrayList<Integer>());
}
}
int[][] show = new int[hight][];
for(int i = 0;i < store.size()-1; i++){
ArrayList<Integer> temp = store.get(i);
System.out.println("size: "+temp.size());
show[i] = new int[temp.size()];
for(int j = 0;j < temp.size(); j++){
show[i][j] = temp.get(j);
System.out.print(temp.get(j)+" ");
}
System.out.println();
}
return show;
}
public static void main(String[] args) {
TreeNode root = new TreeNode(1);
root.left = new TreeNode(2);
root.left.right = new TreeNode(3);
root.left.right.left = new TreeNode(4);
root.left.right.right = new TreeNode(5);
int[][] result = printTree(root);
}
}
请用递归方式实现二叉树的先序、中序和后序的遍历打印。
给定一个二叉树的根结点root，请依次返回二叉树的先序，中序和后续遍历(二维数组的形式)。
public void FirstSearch(TreeNode cur,ArrayList list){
if(cur==null)
return ;
list.add(cur.val);
FirstSearch(cur.left,list);
FirstSearch(cur.right,list);
}
public void MiddleSearch(TreeNode cur,ArrayList list){
if(cur==null)
return ;
MiddleSearch(cur.left,list);
list.add(cur.val);
MiddleSearch(cur.right,list);
}
public void LastSearch(TreeNode cur,ArrayList list){
if(cur==null)
return ;
LastSearch(cur.left,list);
LastSearch(cur.right,list);
list.add(cur.val);
}
public int[][] convert(TreeNode root) {
// write code here
ArrayList<Integer> nodeList = new ArrayList<Integer>();
FirstSearch(root,nodeList);
int[][] result = new int[3][nodeList.size()];
for(int i = 0;i < nodeList.size(); i++)
result[0][i] = nodeList.get(i);
nodeList.clear();
MiddleSearch(root,nodeList);
for(int i = 0;i < nodeList.size(); i++)
result[1][i] = nodeList.get(i);
nodeList.clear();
LastSearch(root,nodeList);
for(int i = 0;i < nodeList.size(); i++)
result[2][i] = nodeList.get(i);
nodeList.clear();
return result;
}
首先我们介绍二叉树先序序列化的方式，假设序列化的结果字符串为str，初始时str等于空字符串。先序遍历二叉树，如果遇到空节点，就在str的末尾加上“#!”，“#”表示这个节点为空，节点值不存在，当然你也可以用其他的特殊字符，“!”表示一个值的结束。如果遇到不为空的节点，假设节点值为3，就在str的末尾加上“3!”。现在请你实现树的先序序列化。
给定树的根结点root，请返回二叉树序列化后的字符串。
public void FristSearch(TreeNode cur,StringBuffer sb){
if(cur==null){
sb.append("#!");
return ;
}
sb.append(cur.val+"!");
FristSearch(cur.left,sb);
FristSearch(cur.right,sb);
}
public String toString(TreeNode root) {
// write code here
StringBuffer sb = new StringBuffer();
FristSearch(root,sb);
return sb.toString();
}
有一棵二叉树，请设计一个算法判断这棵二叉树是否为平衡二叉树。
给定二叉树的根结点root，请返回一个bool值，代表这棵树是否为平衡二叉树。
class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
public int Search(TreeNode cur,int level,boolean[] result){
if(cur==null)
return level;
if(!result[0])
return level;
int lH = Search(cur.left,level+1,result);
if(!result[0])
return level;
int rH = Search(cur.right,level+1,result);
if(Math.abs(lH-rH)>1)
result[0] = false;
return Math.max(lH, rH);
}
public boolean check(TreeNode root) {
// write code here
boolean[] result = new boolean[1];
result[0] = true;
Search(root,1,result);
return result[0];
}
解法2
import java.util.*;
/*
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}*/
public class CheckBalance {
public static boolean check(TreeNode root){
return chk(root)>=0;
}
private static int chk(TreeNode root) {
if (root==null) return 0;
int l=chk(root.left),r=chk(root.right);//左右两边深度
if (l<0||r<0) return -1;
if ((Math.abs(r-l)>1))return -1;//当左右两边深度差超过1时 返回-1
return r>l?r+1:l+1;
}
}
一棵二叉树原本是搜索二叉树，但是其中有两个节点调换了位置，使得这棵二叉树不再是搜索二叉树，请找到这两个错误节点并返回他们的值。保证二叉树中结点的值各不相同。
给定一棵树的根结点，请返回两个调换了位置的值，其中小的值在前。
核心代码：
public int[] findError(TreeNode root) {
// write code here
int res[]=new int[2];
TreeNode err[]=new TreeNode[2];
Stack<TreeNode> st=new Stack<TreeNode>();
TreeNode pre=null;
while(!st.isEmpty()||root!=null){
if(root!=null){
st.push(root);
root=root.left;
}else{
root=st.pop();
if(pre!=null&&pre.val>root.val){
err[0]=err[0]==null?pre:err[0];
err[1]=root;
}
pre=root;
root=root.right;
}
}
res[1]=err[0].val;
res[0]=err[1].val;
return res;
}
解法二:
public class FindErrorNode {
public int[] findError(TreeNode root) {
// write code here
ArrayList<Integer> list = new ArrayList<Integer>();
int[] result = new int[2];
LastSearch(root,list);
int count = 0;
for(int i = 0;i < list.size()-1; i++)
if(list.get(i)>list.get(i+1)){
result[1] = list.get(i+1);
if(count==0){
result[0] = list.get(i);
//result[1] = list.get(i+1)
}
count++;
}
result[0] = result[0]^result[1];
result[1] = result[0]^result[1];
result[0] = result[0]^result[1];
return result;
}
private void LastSearch(TreeNode cur,List list){
if(cur==null){
return ;
}
LastSearch(cur.left,list);
list.add(cur.val);
LastSearch(cur.right,list);
}
}
请用非递归方式实现二叉树的先序、中序和后序的遍历打印。
给定一个二叉树的根结点root，请依次返回二叉树的先序，中序和后续遍历(二维数组的形式)。
package com.oj;
import java.util.ArrayList;
import java.util.Stack;
class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
public class TreeSearch {
public void PreSearch(TreeNode root,ArrayList<Integer> list){
Stack stack = new Stack();
stack.push(root);
while(!stack.isEmpty()){
TreeNode temp = (TreeNode) stack.pop();
list.add(temp.val);
if(temp.right!=null)
stack.push(temp.right);
if(temp.left!=null)
stack.push(temp.left);
}
}
public void MidSearch(TreeNode root,ArrayList<Integer> list){
Stack stack = new Stack();
TreeNode cur = root;
stack.push(cur);
cur = cur.left;
while(!stack.isEmpty()||cur!=null){
if(cur!=null){
stack.push(cur);
cur = cur.left;
}else{
cur = (TreeNode) stack.pop();
list.add(cur.val);
cur = cur.right;
}
}
}
public void LastSearch(TreeNode root,ArrayList<Integer> list){
Stack s1 = new Stack();
Stack s2 = new Stack();
TreeNode cur = root;
s1.push(cur);
while(!s1.isEmpty()){
cur = (TreeNode) s1.pop();
s2.push(cur);
if(cur.left!=null){
s1.push(cur.left);
}
if(cur.right!=null){
s1.push(cur.right);
}
}
while(!s2.isEmpty()){
TreeNode temp = (TreeNode) s2.pop();
list.add(temp.val);
}
}
public int[][] convert(TreeNode root) {
// write code here
int[][] result = new int[3][];
ArrayList<Integer> list = new ArrayList<Integer>();
PreSearch(root,list);
result[0] = new int[list.size()];
for(int i = 0;i < list.size(); i++){
result[0][i] = list.get(i);
}
list.clear();
MidSearch(root,list);
result[1] = new int[list.size()];
for(int i = 0;i < list.size(); i++){
result[1][i] = list.get(i);
}
list.clear();
LastSearch(root,list);
result[2] = new int[list.size()];
for(int i = 0;i < list.size(); i++){
result[2][i] = list.get(i);
}
return result;
}
}
从二叉树的节点A出发，可以向上或者向下走，但沿途的节点只能经过一次，当到达节点B时，路径上的节点数叫作A到B的距离。对于给定的一棵二叉树，求整棵树上节点间的最大距离。
给定一个二叉树的头结点root，请返回最大距离。保证点数大于等于2小于等于500.
import java.util.*;
/*
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}*/
public class LongestDistance {
public int findLongest(TreeNode root) {
int[]record=new int[1];
return process(root,record);
}
public int process(TreeNode root,int[]record){
if(root==null){
record[0]=0;
return 0;
}
int lmax=process(root.left,record);
int maxfromleft=record[0];
int rmax=process(root.right,record);
int maxfromright=record[0];
int cur=maxfromleft+maxfromright+1;
record[0]=Math.max(maxfromleft,maxfromright)+1;
return Math.max(Math.max(lmax,rmax),cur);
}
}
题目描述：重建二叉树
输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode ConstructTree(int[] pre,int[] in,int prestart,int preend,int instart,int inend) throws Exception
{
int rootvalue = pre[prestart];
TreeNode root = new TreeNode(rootvalue);
root.left = null;
root.right = null;
if(prestart==preend){
if(instart==inend&&pre[prestart]==in[instart])
return root;
else
throw new Exception("wrong");
}
int posinorder = instart;
while(posinorder<=inend&&in[posinorder]!=rootvalue)
++posinorder;
if(posinorder==inend&&in[posinorder]!=rootvalue)
throw new Exception("wrong");
int leftLength = posinorder-instart;
int leftPreend = prestart+leftLength;
if(leftLength>0)
{
root.left = ConstructTree(pre,in,prestart+1,leftPreend,instart,posinorder-1);
}
if(leftLength<preend-prestart)
{
root.right = ConstructTree(pre,in,leftPreend+1,preend,posinorder+1,inend);
}
return root;
}
public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
int prestart = 0, preend = pre.length-1;
int instart = 0, inend = in.length-1;
if(preend==0||inend==0)
return null;
try {
return ConstructTree(pre,in,prestart,preend,instart,inend);
} catch (Exception e) {
//e.printStackTrace();
return null;
}
}
}
态度决定高度，细节决定成败，
