搞定面试中的二叉树题目(Java实现)

╰半橙微兮° 2022-05-24 12:46 220阅读 0赞

**本文转载自：https://www.jianshu.com/p/0190985635eb**

这是一篇关于二叉树的文章，总结了二叉树数据结构和算法的相关题目。  
先上二叉树的数据结构：

class TreeNode{
           
        int val;
        //左孩子
        TreeNode left;
        //右孩子
        TreeNode right;
    }

二叉树的题目普遍可以用递归和迭代的方式来解

### 1.求二叉树的最大深度 ###

int maxDeath(TreeNode node){
        if(node==null){
            return 0;
        }
        int left = maxDeath(node.left);
        int right = maxDeath(node.right);
        return Math.max(left,right) + 1;
    }

### 2.求二叉树的最小深度 ###

int getMinDepth(TreeNode root){
            if(root == null){
                return 0;
            }
            return getMin(root);
        }
        int getMin(TreeNode root){
            if(root == null){
                return Integer.MAX_VALUE;
            }
            if(root.left == null&&root.right == null){
                return 1;
            }
            return Math.min(getMin(root.left),getMin(root.right)) + 1;
        }

### 3,求二叉树中节点的个数 ###

int numOfTreeNode(TreeNode root){
            if(root == null){
                return 0;
                
            }
            int left = numOfTreeNode(root.left);
            int right = numOfTreeNode(root.right);
            return left + right + 1;
        }

### 4,求二叉树中叶子节点的个数 ###

int numsOfNoChildNode(TreeNode root){
            if(root == null){
                return 0;
            }
            if(root.left==null&&root.right==null){
                return 1;
            }
            return numsOfNodeTreeNode(root.left)+numsOfNodeTreeNode(root.right);
            
        }

### 5.求二叉树中第k层节点的个数 ###

int numsOfkLevelTreeNode(TreeNode root,int k){
                if(root == null||k<1){
                    return 0;
                }
                if(k==1){
                    return 1;
                }
                int numsLeft = numsOfkLevelTreeNode(root.left,k-1);
                int numsRight = numsOfkLevelTreeNode(root.right,k-1);
                return numsLeft + numsRight;
            }

### 6.判断二叉树是否是平衡二叉树 ###

boolean isBalanced(TreeNode node){
            return maxDeath2(node)!=-1;
        }
        int maxDeath2(TreeNode node){
            if(node == null){
                return 0;
            }
            int left = maxDeath2(node.left);
            int right = maxDeath2(node.right);
            if(left==-1||right==-1||Math.abs(left-right)>1){
                return -1;
            }
            return Math.max(left, right) + 1;
        }

### 7.判断二叉树是否是完全二叉树 ###

> 什么是完全二叉树呢？[参见][Link 1]

boolean isCompleteTreeNode(TreeNode root){
            if(root == null){
                return false;
            }
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.add(root);
            boolean result = true;
            boolean hasNoChild = false;
            while(!queue.isEmpty()){
                TreeNode current = queue.remove();
                if(hasNoChild){
                    if(current.left!=null||current.right!=null){
                        result = false;
                        break;
                    }
                }else{
                    if(current.left!=null&&current.right!=null){
                        queue.add(current.left);
                        queue.add(current.right);
                    }else if(current.left!=null&&current.right==null){
                        queue.add(current.left);
                        hasNoChild = true;
                        
                    }else if(current.left==null&&current.right!=null){
                        result = false;
                        break;
                    }else{
                        hasNoChild = true;
                    }
                }
                
            }
            return result;
        }

### 8.两个二叉树是否完全相同 ###

boolean isSameTreeNode(TreeNode t1,TreeNode t2){
            if(t1==null&&t2==null){
                return true;
            }
            else if(t1==null||t2==null){
                return false;
            }
            if(t1.val != t2.val){
                return false;
            }
            boolean left = isSameTreeNode(t1.left,t2.left);
            boolean right = isSameTreeNode(t1.right,t2.right);
            return left&&right;
            
        }

### 9.两个二叉树是否互为镜像 ###

boolean isMirror(TreeNode t1,TreeNode t2){
            if(t1==null&&t2==null){
                return true;
            }
            if(t1==null||t2==null){
                return false;
            }
            if(t1.val != t2.val){
                return false;
            }
            return isMirror(t1.left,t2.right)&&isMirror(t1.right,t2.left);
            
        }

### 10.翻转二叉树or镜像二叉树 ###

TreeNode mirrorTreeNode(TreeNode root){
            if(root == null){
                return null;
            }
            TreeNode left = mirrorTreeNode(root.left);
            TreeNode right = mirrorTreeNode(root.right);
            root.left = right;
            root.right = left;
            return root;
        }

### 11.求两个二叉树的最低公共祖先节点 ###

TreeNode getLastCommonParent(TreeNode root,TreeNode t1,TreeNode t2){
            if(findNode(root.left,t1)){
                if(findNode(root.right,t2)){
                    return root;
                }else{
                    return getLastCommonParent(root.left,t1,t2);
                }
            }else{
                if(findNode(root.left,t2)){
                    return root;
                }else{
                    return getLastCommonParent(root.right,t1,t2)
                }
            }
        }
        // 查找节点node是否在当前 二叉树中
        boolean findNode(TreeNode root,TreeNode node){
            if(root == null || node == null){
                return false;
            }
            if(root == node){
                return true;
            }
            boolean found = findNode(root.left,node);
            if(!found){
                found = findNode(root.right,node);
            }
            return found;
        }

### 12.二叉树的前序遍历 ###

#### 迭代解法 ####

ArrayList<Integer> preOrder(TreeNode root){
            Stack<TreeNode> stack = new Stack<TreeNode>();
            ArrayList<Integer> list = new ArrayList<Integer>();
            if(root == null){
                return list;
            }
            stack.push(root);
            while(!stack.empty()){
                TreeNode node = stack.pop();
                list.add(node.val);
                if(node.right!=null){
                    stack.push(node.right);
                }
                if(node.left != null){
                    stack.push(node.left);
                }
                
            }
            return list;
        }

#### 递归解法 ####

ArrayList<Integer> preOrderReverse(TreeNode root){
            ArrayList<Integer> result = new ArrayList<Integer>();
            preOrder2(root,result);
            return result;
            
        }
        void preOrder2(TreeNode root,ArrayList<Integer> result){
            if(root == null){
                return;
            }
            result.add(root.val);
            preOrder2(root.left,result);
            preOrder2(root.right,result);
        }

### 13.二叉树的中序遍历 ###

ArrayList<Integer> inOrder(TreeNode root){
            ArrayList<Integer> list = new ArrayList<<Integer>();
            Stack<TreeNode> stack = new Stack<TreeNode>();
            TreeNode current = root;
            while(current != null|| !stack.empty()){
                while(current != null){
                    stack.add(current);
                    current = current.left;
                }
                current = stack.peek();
                stack.pop();
                list.add(current.val);
                current = current.right;
                
            }
            return list;
            
        }

### 14.二叉树的后序遍历 ###

ArrayList<Integer> postOrder(TreeNode root){
            ArrayList<Integer> list = new ArrayList<Integer>();
            if(root == null){
                return list;
            }
            list.addAll(postOrder(root.left));
            list.addAll(postOrder(root.right));
            list.add(root.val);
            return list;
        }

### 15.前序遍历和后序遍历构造二叉树 ###

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

### 16.在二叉树中插入节点 ###

TreeNode insertNode(TreeNode root,TreeNode node){
            if(root == node){
                return node;
            }
            TreeNode tmp = new TreeNode();
            tmp = root;
            TreeNode last = null;
            while(tmp!=null){
                last = tmp;
                if(tmp.val>node.val){
                    tmp = tmp.left;
                }else{
                    tmp = tmp.right;
                }
            }
            if(last!=null){
                if(last.val>node.val){
                    last.left = node;
                }else{
                    last.right = node;
                }
            }
            return root;
        }

### 17.输入一个二叉树和一个整数，打印出二叉树中节点值的和等于输入整数所有的路径 ###

void findPath(TreeNode r,int i){
            if(root == null){
                return;
            }
            Stack<Integer> stack = new Stack<Integer>();
            int currentSum = 0;
            findPath(r, i, stack, currentSum);
            
        }
        void findPath(TreeNode r,int i,Stack<Integer> stack,int currentSum){
            currentSum+=r.val;
            stack.push(r.val);
            if(r.left==null&&r.right==null){
                if(currentSum==i){
                    for(int path:stack){
                        System.out.println(path);
                    }
                    
                }
            }
            if(r.left!=null){
                findPath(r.left, i, stack, currentSum);
            }
            if(r.right!=null){
                findPath(r.right, i, stack, currentSum);
            }
            stack.pop();
        }

### 18.二叉树的搜索区间 ###

> 给定两个值 k1 和 k2（k1 < k2）和一个二叉查找树的根节点。找到树中所有值在 k1 到 k2 范围内的节点。即打印所有x (k1 <= x <= k2) 其中 x 是二叉查找树的中的节点值。返回所有升序的节点值。

ArrayList<Integer> result;
        ArrayList<Integer> searchRange(TreeNode root,int k1,int k2){
            result = new ArrayList<Integer>();
            searchHelper(root,k1,k2);
            return result;
        }
        void searchHelper(TreeNode root,int k1,int k2){
            if(root == null){
                return;
            }
            if(root.val>k1){
                searchHelper(root.left,k1,k2);
            }
            if(root.val>=k1&&root.val<=k2){
                result.add(root.val);
            }
            if(root.val<k2){
                searchHelper(root.right,k1,k2);
            }
        }

### 19.二叉树的层次遍历 ###

ArrayList<ArrayList<Integer>> levelOrder(TreeNode root){
            ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
            if(root == null){
                return result;
            }
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            while(!queue.isEmpty()){
                int size = queue.size();
                ArrayList<<Integer> level = new ArrayList<Integer>():
                for(int i = 0;i < size ;i++){
                    TreeNode node = queue.poll();
                    level.add(node.val);
                    if(node.left != null){
                        queue.offer(node.left);
                    }
                    if(node.right != null){
                        queue.offer(node.right);
                    }
                } 
                result.add(Level);
            }
            return result;
        }

### 20.二叉树内两个节点的最长距离 ###

> 二叉树中两个节点的最长距离可能有三种情况：  
> 1.左子树的最大深度+右子树的最大深度为二叉树的最长距离  
> 2.左子树中的最长距离即为二叉树的最长距离  
> 3.右子树种的最长距离即为二叉树的最长距离  
> 因此，递归求解即可

private static class Result{  
        int maxDistance;  
        int maxDepth;  
        public Result() {  
        }  
      
        public Result(int maxDistance, int maxDepth) {  
            this.maxDistance = maxDistance;  
            this.maxDepth = maxDepth;  
        }  
    }  
        int getMaxDistance(TreeNode root){
          return getMaxDistanceResult(root).maxDistance;
        }
        Result getMaxDistanceResult(TreeNode root){
            if(root == null){
                Result empty = new Result(0,-1);
                return empty;
            }
            Result lmd = getMaxDistanceResult(root.left);
            Result rmd = getMaxDistanceResult(root.right);
            Result result = new Result();
            result.maxDepth = Math.max(lmd.maxDepth,rmd.maxDepth) + 1;
            result.maxDistance = Math.max(lmd.maxDepth + rmd.maxDepth,Math.max(lmd.maxDistance,rmd.maxDistance));
            return result;
        }

### 21.不同的二叉树 ###

> 给出 *n*，问由 1...*n* 为节点组成的不同的二叉查找树有多少种？

int numTrees(int n ){
            int[] counts = new int[n+2];
            counts[0] = 1;
            counts[1] = 1;
            for(int i = 2;i<=n;i++){
                for(int j = 0;j<i;j++){
                    counts[i] += counts[j] * counts[i-j-1];
                }
            }
            return counts[n];
        }

### 22.判断二叉树是否是合法的二叉查找树(BST) ###

> 一棵BST定义为：  
> 节点的左子树中的值要**严格**小于该节点的值。  
> 节点的右子树中的值要**严格**大于该节点的值。  
> 左右子树也必须是二叉查找树。  
> 一个节点的树也是二叉查找树。

public int lastVal = Integer.MAX_VALUE;
        public boolean firstNode = true;
        public boolean isValidBST(TreeNode root) {
            // write your code here
            if(root==null){
                return true;
            }
            if(!isValidBST(root.left)){
                return false;
            }
            if(!firstNode&&lastVal >= root.val){
                return false;
            }
            firstNode = false;
            lastVal = root.val;
            if (!isValidBST(root.right)) {
                return false;
            }
            return true;
        }

作者：IOExceptioner  
链接：https://www.jianshu.com/p/0190985635eb  
來源：简书  
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

[Link 1]: https://link.jianshu.com?t=http://baike.baidu.com/link?url=eSO2K_nYvOs22ts7GHo2g3KK80U8XvDuZeez4PecVIzqkE_bZqTipNg9khef-WHjMBfKSgu5EVwRie_kG5MhIbqWpFNx1A3pvdV926Rnsd-Q4H7VXgqiuG71B7ygxl4tOsKHIWkN41OGFNKQj97jYq