文章目录
- 一、二叉树前中后遍历
- 二、获取节点个数
- 三.获取叶子节点个数
- 四.获取第k层节点个数
- 五.求二叉树的高度，时间复杂度O（N）
- 六.检测值为value的元素是否存在
- [七. 检查两颗树是否相同](#七. 检查两颗树是否相同)
- 八.判断一棵二叉树是不是平衡二叉树
- [九.一个二叉树的根节点 root ， 检查它是否轴对称](#九.一个二叉树的根节点 root ， 检查它是否轴对称)
- [十. 判断subRoot是不是root的子树](#十. 判断subRoot是不是root的子树)
- 十一.翻转二叉树
- 总结
一、二叉树前中后遍历
java
public class BinaryTree {
static class TreeNode{
public char val;
public TreeNode left;
public TreeNode right;
public TreeNode(char val) {
this.val = val;
}
}
public TreeNode creatTree(){
TreeNode A = new TreeNode('A');
TreeNode B = new TreeNode('B');
TreeNode C = new TreeNode('C');
TreeNode D = new TreeNode('D');
TreeNode E = new TreeNode('E');
TreeNode F = new TreeNode('F');
TreeNode G = new TreeNode('G');
TreeNode H = new TreeNode('H');
A.left = B;
A.right = C;
B.left = D;
B.right = E;
C.left = F;
C.right = G;
E.right = H;
return A;
}
//前序遍历
void preOrder(TreeNode root){
//根左右
if(root == null){
return;
}
System.out.print(root.val+" ");
preOrder(root.left);
preOrder(root.right);
}
//中序遍历
void inOrder(TreeNode root){
//左根右
if(root == null){
return;
}
inOrder(root.left);
System.out.print(root.val+" ");
inOrder(root.right);
}
//后序遍历
void postOrder(TreeNode root){
//左右根
if(root == null){
return;
}
postOrder(root.left);
postOrder(root.right);
System.out.print(root.val+" ");
}
public static void main(String[] args) {
BinaryTree binaryTree = new BinaryTree();
BinaryTree.TreeNode root = binaryTree.creatTree();
binaryTree.preOrder(root);
System.out.println();
binaryTree.inOrder(root);
System.out.println();
binaryTree.postOrder(root);
}
}二、获取节点个数
java
public int nodeSize;
int size(TreeNode root){
if (root == null){
return 0;
}
nodeSize++;
size(root.left);
size(root.right);
return nodeSize;
}
//子问题思路解决：整棵树的节点=左子树的节点+右子数的节点+1
int size2(TreeNode root){
if (root == null){
return 0;
}
return size2(root.left)+size2(root.right)+1;
}三.获取叶子节点个数
java
public int leafSize;
int getLeafNodeCount(TreeNode root){
if (root == null){
return 0;
}
if (root.left == null && root.right == null){
leafSize++;
}
getLeafNodeCount(root.left);
getLeafNodeCount(root.right);
return leafSize;
}四.获取第k层节点个数
java
public int leveSize;
int getleveNodeCount(TreeNode root,int k){
if(root == null){
return 0;
}
if (k == 1) {
return 1;
}
return getleveNodeCount(root.left,k-1)
+ getleveNodeCount(root.right,k-1);
}五.求二叉树的高度，时间复杂度O（N）
java
int getHeight(TreeNode root){
if (root == null){
return 0;
}
int leftHeight = getHeight(root.left);
int rightHeight = getHeight(root.right);
return leftHeight > rightHeight?leftHeight+1:rightHeight+1;
}六.检测值为value的元素是否存在
java
TreeNode find(TreeNode root,int val){
if (root == null){
return null;
}
if (root.val == val){
return root;
}
TreeNode ret1 = find(root.left,val);
if (ret1 != null){
return ret1;
}
TreeNode ret2 = find(root.right,val);
if (ret2 != null){
return ret2;
}
return null;
}七. 检查两颗树是否相同
java
public boolean isSameTree(TreeNode p, TreeNode q) {
if( (p == null && q !=null) || (p != null && q == null) ){
return false;
}
if(p == null && q == null){
return true;
}
if(p.val !=q.val){
return false;
}
return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);
}八.判断一棵二叉树是不是平衡二叉树
时间复杂度O（N）
java
public boolean isBalanced(TreeNode root) {
if(root == null){
return true;
}
return getHeight(root)>=0;
}
int getHeight(TreeNode root) {
if (root == null){
return 0;
}
int leftHeight =getHeight(root.left);
if(leftHeight < 0){
return -1;
}
int rightHeight = getHeight(root.right);
if(leftHeight >= 0 && rightHeight >= 0 && Math.abs(leftHeight - rightHeight) <=1){
return Math.max(leftHeight,rightHeight)+1;
}else{
return -1;
}
}九.一个二叉树的根节点 root ， 检查它是否轴对称
java
public boolean isSymmetric(TreeNode root) {
if(root == null) return true;
return isSymmetricChild(root.left,root.right);
}
private boolean isSymmetricChild(TreeNode leftTree,TreeNode rightTree){
if((leftTree == null && rightTree != null) ||(leftTree != null && rightTree == null)){
return false;
}
if(leftTree == null && rightTree == null){
return true;
}
if(leftTree.val != rightTree.val){
return false;
}
return isSymmetricChild(leftTree.left,rightTree.right) && isSymmetricChild(leftTree.right,rightTree.left);
}十. 判断subRoot是不是root的子树
java
//时间复杂度：O(m*n)
public boolean isSubtree(TreeNode root,TreeNode subTree) {
if(root == null || subTree == null) {
return false;
}
//判断根节点是否相同
if (isSameTree(root,subTree)){
return true;
}
//判断左子树是否相同
if(isSubtree(root.left, subTree)) {
return true;
}
//判断右子树是否相同
if(isSubtree(root.right, subTree)) {
return true;
}
return false;
}
public boolean isSameTree(TreeNode p, TreeNode q) {
if( (p == null && q !=null) || (p != null && q == null) ){
return false;
}
if(p == null && q == null){
return true;
}
if(p.val !=q.val){
return false;
}
return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);
}十一.翻转二叉树
java
public TreeNode invertTree(TreeNode root){
if(root == null){
return null;
}
if(root.left == null && root.right == null){
return root;
}
//交换左右节点
TreeNode tmp = root.left;
root.left = root.right;
root.right = tmp;
//交换左树
invertTree(root.left);
//交换右树
invertTree(root.right);
return root;
}总结
今天复习二叉树练习。
时隔一个月继续更新文章，学校的考试周太可怕了=.= 怕挂科所以全身心投入复习，平时都学编程了，学校的课靠最后几周学完全部，也是挺厉害的哈哈。