知识改变命运数据结构【二叉树OJ题】

1. 检查两颗树是否相同OJ链接

 class Solution {
        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;
            }
            boolean left=isSameTree(p.left,q.left);
            boolean right=isSameTree(p.right,q.right);
            if (left&&right) {
                return true;
            } else {
                return false;
            }

        }
 }

2. 另一颗树的子树。OJ链接

class Solution {
   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);
        }
        public boolean isSubtree(TreeNode root, TreeNode subRoot) {
            if (root==null&&subRoot!=null) {
                return false;
            }
            if (root==null&&subRoot==null) {
                return true;
            }
            if(isSameTree(root,subRoot)) {
                return true;
            }
            if(isSubtree(root.left,subRoot)) {
                return true;
            }
            if (isSubtree(root.right,subRoot)) {
                return true;
            }
            return false;
        }
}

3. 翻转二叉树。Oj链接

class Solution {
  public TreeNode invertTree(TreeNode root) {
        if(root==null) {
            return null;
        }
        TreeNode tem=root.left;
        root.left=root.right;
        root.right=tem;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}

4. 判断一颗二叉树是否是平衡二叉树。OJ链接

class Solution {
      public boolean isBalanced(TreeNode root) {
        if(root==null) {
            return true;
        }
        if(Math.abs((getHeight(root.left)-getHeight(root.right)))>1) {
            return false;
        }
        return isBalanced(root.left)&&isBalanced(root.right);
    }
     int getHeight(TreeNode root) {
        if(root==null) {
            return 0;
        }
        int leftHeight=getHeight(root.left);
        int rightHeight=getHeight(root.right);
        return Math.max(leftHeight+1,rightHeight+1);
    }
}

5. 对称二叉树。OJ链接

class Solution {
      public boolean isBalanced(TreeNode root) {
        if(root==null) {
            return true;
        }
        if(Math.abs((getHeight(root.left)-getHeight(root.right)))>1) {
            return false;
        }
        return isBalanced(root.left)&&isBalanced(root.right);
    }
     int getHeight(TreeNode root) {
        if(root==null) {
            return 0;
        }
        int leftHeight=getHeight(root.left);
        int rightHeight=getHeight(root.right);
        return Math.max(leftHeight+1,rightHeight+1);
    }
}

6. 二叉树的构建及遍历。OJ链接

public class Main {
    static class TreeNode {
        char val;
        TreeNode left;
        TreeNode right;
        public TreeNode(char val) {
            this.val = val;
        }
    }
    public static int i = 0;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNextLine()) { // 注意 while 处理多个 case
            String ch = in.nextLine();
            TreeNode root = creatTree(ch);
            inOrder(root);
        }
    }
    public static TreeNode creatTree(String ch) {
        TreeNode root = null;

        char ch2 = ch.charAt(i);
        if (ch2 != '#') {
            root = new TreeNode(ch2);
            i++;
            root.left = creatTree(ch);
            root.right = creatTree(ch);
        } else {
            i++;
        }
        return root;
    }
    public static void  inOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        inOrder(root.left);
        System.out.print(root.val + " ");
        inOrder(root.right);
    }
}

7. 二叉树的分层遍历。OJ链接

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List <List<Integer>> ret =new LinkedList<>();
        if(root==null) {
            return ret;
        }
        Queue<TreeNode> queue=new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            List<Integer> list=new ArrayList<>();
            int size=queue.size();
            while(size!=0) {
                TreeNode cur=queue.poll();
                size--;
                list.add(cur.val);
                if(cur.left!=null) {
                    queue.offer(cur.left);
                }
                if(cur.right!=null) {
                    queue.offer(cur.right);
                }
            }
            ret.add(list);
        }
        return ret;
    }
}

8. 给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。OJ链接

方法1：

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root==null) {
            return null;
        }
        if(root==p) {
            return p;
        }
        if(root==q) {
            return q;
        }
        TreeNode leftRoot=lowestCommonAncestor(root.left,p,q);
        TreeNode rightRoot=lowestCommonAncestor(root.right,p,q);
        if(leftRoot!=null&&rightRoot!=null) {
            return root;
        }
        if(leftRoot==null) {
            return rightRoot;
        }
        if(rightRoot==null) {
            return leftRoot;
        }
        return null;
    }
}

9. 根据一棵树的前序遍历与中序遍历构造二叉树。 OJ链接

class Solution {
   public int i=0;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        TreeNode root;
        root=greatTreeSon(preorder,inorder,0,inorder.length-1);
        return root;
    }
    public TreeNode greatTreeSon(int[] preorder, int[] inorder0,int inBegin,int inEnd) {
        if(inBegin>inEnd) {
            return null;
        }
        TreeNode root=new TreeNode(preorder[i]);
        int index=findKey(preorder[i],inorder0,inBegin,inEnd);
        i++;
        root.left=greatTreeSon(preorder,inorder0,inBegin,index-1);
        root.right=greatTreeSon(preorder,inorder0,index+1,inEnd);
        return root;
        
    }
    private int findKey(int key, int[] inorder0,int inBegin,int inEnd) {
        for (int j = inBegin; j <=inEnd ; j++) {
            if(inorder0[j]==key) {
                return j;
            }
        }
        return -1;
    }
}

10. 根据一棵树的中序遍历与后序遍历构造二叉树。OJ链接

11. 二叉树创建字符串。OJ链接

class Solution {
    public String tree2str(TreeNode root) {
        StringBuilder stringBuilder=new StringBuilder();
        GreatString(stringBuilder,root);
        return stringBuilder.toString();
    }
    public void GreatString(StringBuilder stringBuilder,TreeNode root) {
        if(root==null) {
            return;
        }
        stringBuilder.append(root.val);
        if (root.left!=null) {
            stringBuilder.append("(");
            GreatString(stringBuilder,root.left);
            stringBuilder.append(")");
        } else {
            if(root.right==null) {
                return;
            } else {
                stringBuilder.append("()");
            }
        }
        if(root.right!=null) {
            stringBuilder.append("(");
            GreatString(stringBuilder,root.right);
            stringBuilder.append(")");
        } else {
            return;
        }
    }
}