当前位置 博文首页 > 强仔不强的博客:龙生九子,各有不同,看这四种二叉树如何各显神
对于高度为k的,有n个节点的二叉树,当且仅当其每个节点都与高度为k的满二叉树中编号从1~n的节点一一对应时称为完全二叉树
需要用到队列
public boolean isCompleteTree2(Node root){
if(root == null){
return true;
}
Queue<Node> queue = new LinkedList<>();
queue.offer(root);
while(true) {
Node top = queue.poll();
if (top == null) {
while(!queue.isEmpty()) {
if(queue.poll() != null){
return false;
}
}
return true;
} else {
queue.offer(top.left);
queue.offer(top.right);
}
}
}
每个节点的左树和右树的高度差<=1
如果一棵树为平衡二叉树,那么这棵树的每棵子树都是平衡二叉树
public int Hight(Node root){
if(root == null){
return 0;
}
int left = Hight(root.left);
int right = Hight(root.right);
return ((left > right) ? left : right) +1;
}
public boolean isBalanced(Node root){
if(root == null){
return true;
}
int leftHight = Hight(root.left);
int rightHIght = Hight(root.right);
if(Math.abs(leftHight - rightHIght) >1 ){
return false;
}
return isBalanced(root.left) && isBalanced(root.right);
}
此代码时间复杂度O(n^2)
下面这个代码的时间复杂度为O(n^2)
即在求二叉树高度的时候就判断是否是平衡二叉树
root为E的方法中,leftHight = 0,rightHight = 2,直接返回 -1;
即root为B的方法中,leftHight = 1;rightHight = -1;直接返回-1;
…
public int Hight(Node root){
if(root == null){
return 0;
}
int leftHight = Hight(root.left);
int rightHight = Hight(root.right);
if(leftHight == -1 || rightHight == -1 || Math.abs(leftHight - rightHight) > 1){
return -1;
}
return (leftHight > rightHight ? leftHight : rightHight) + 1;
}
public boolean isBalanced(Node root){
if(Hight(root) != -1){
return true;
}
return false;
}
对称二叉树
非对称二叉树
public boolean isSymmeticChild(Node leftroot , Node rightroot){
if(leftroot == null && rightroot == null){
return true;
}
if(leftroot == null || rightroot == null){
return false;
}
if(leftroot.getVal() != rightroot.getVal()){
return false;
}
return isSymmeticChild(leftroot.right , rightroot.left) && isSymmeticChild(leftroot.left , rightroot.right);
}
public boolean isSymmetric(Node root){
if(root == null){
return true;
}
return isSymmeticChild(root.left,root.right);
}
写出一个二叉树的镜像二叉树
public Node Mirror (Node root){
if(root == null){
return root;
}
if(root.left == null && root.right == null){
return root;
}
Node temp = root.left;
root.left = root.right;
root.right = temp;
root.left = Mirror(root.left);
root.left = Mirror(root.right);
return root;
}
cs