JAVA二叉树相关习题5

1. 二叉树前序非递归遍历实现。

. - 力扣（LeetCode）

递归的实现

public List<TreeNode> preOrder1(TreeNode root){
        List<TreeNode> ret=new ArrayList<>();
        if(root == null)
            return ret;
        ret.add(root);
        List<TreeNode> leftTree = preOrder1(root.left);
        ret.addAll(leftTree);
        List<TreeNode> rightTree = preOrder1(root.right);
        ret.addAll(rightTree);
        return ret;
    }

与栈相结合

public void preOrderNor(TreeNode root) {
        if(root == null) return;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode top = null;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                System.out.print(cur.val + " ");
                cur = cur.left;
            }

            top = stack.pop();
            cur = top.right;
        }
    }

2. 二叉树中序非递归遍历实现。

. - 力扣（LeetCode）

3. 二叉树后序非递归遍历实现。

. - 力扣（LeetCode）

4. 检查两颗树是否相同。

. - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        //一个为空一个不为空
        if((p==null&&q!=null)||(p!=null&&q==null)){
            return false;
        }
        //两个都为空
        if(p==null&&q==null){
            return true;
        }

        //两个都不为空
        if(p.val!=q.val){
            return false;
        }
        return isSameTree(p.right, q.right)&&isSameTree(p.left, q.left);
    }
}

5. 另一颗树的子树。

. - 力扣（LeetCode）

//判断左数和其是否相同

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if(p == null && q != null || p != null && q == null ) {
            return false;
        }
        //两个都为空呢？
        if(p == null && q == null) {
            return true;
        }
        //一定是两个引用 都不为空 | 两个都不为空呢？
        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){
            return false;
        }
        if(isSameTree(root,subRoot)){
            return true;
        }
        if(isSubtree(root.left,subRoot)){
            return true;
        }
        if(isSubtree(root.right,subRoot)){
            return true;
        }
        return false;
    }
}

6.翻转二叉树。

. - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if(root==null){
            return null;
        }
        TreeNode tmp=root.left;
        root.left=root.right;
        root.right=tmp;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}

7. 判断一颗二叉树是否是平衡二叉树。

但是上述方式复杂度较高

上述代码速度很慢！！！

. - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int getHeight(TreeNode root) {
            if(root == null)
                return 0;
            int leftHeight = getHeight(root.left);
            int rightHeight = getHeight(root.right);
            if(leftHeight>=0&&rightHeight>=0&&Math.abs(leftHeight-rightHeight)<=1){
                return Math.max(leftHeight,rightHeight)+1;
            }else{
                return -1;
            }
            
        }
    public boolean isBalanced(TreeNode root) {
        if(root==null){
            return true;
        }
        return getHeight(root)>=0;
    }
}

8.对称二叉树。

. - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null){
            return true;
        }
        return isSymmetricChild(root.left,root.right);
    }
    public boolean isSymmetricChild(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 isSymmetricChild(p.left,q.right)&&isSymmetricChild(p.right,q.left);
    }
}

9.二叉树的构建及遍历。

二叉树遍历_牛客题霸_牛客网

hasNext、hasNextLine、next、nextLine保姆级详解-CSDN博客

import java.util.Scanner;

//从头到尾进行书写
class TreeNode{
    char val;
    TreeNode left;
    TreeNode right;
    //无参构造
    TreeNode(){

    }
    //有参构造
    TreeNode(char val){
        this.val=val;
    }
}
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    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 str=in.nextLine();
            TreeNode root=createTree(str);
            inOrder(root);
        }
    }

    public static int i=0;
    public static TreeNode createTree(String str){
        TreeNode root =null;
        if(str.charAt(i)!='#'){
            root=new TreeNode(str.charAt(i));
            i++;
            root.left=createTree(str);
            root.right=createTree(str);
            
        }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);
    }
}

10.二叉树的分层遍历

. - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ret = new ArrayList<>();
        if(root == null) {
            return ret;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            //求一下当前队列的大小  4
            int size = queue.size();//4
            List<Integer> tmp = new ArrayList<>();
            while (size != 0) {
                // 出队列4次 相当于把 这一层的节点都 出队了
                TreeNode cur = queue.poll();
                //System.out.print(cur.val+" ");
                tmp.add(cur.val);
                size--;//0
                if(cur.left != null) {
                    queue.offer(cur.left);
                }
                if(cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            ret.add(tmp);
        }
        return ret;
    }
}

11.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。

. - 力扣（LeetCode）

方法一：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null) {
            return null;
        }
        if(root == p || root == q) {
            return root;
        }
        TreeNode leftTree = lowestCommonAncestor(root.left,p,q);
        TreeNode rightTree = lowestCommonAncestor(root.right,p,q);
        if(leftTree != null && rightTree != null) {
            return root;
        }else if(leftTree != null) {
            return leftTree;
        }else {
            return rightTree;
        }
    }
}

方法二：与栈相结合

/**
     * 找到root 到 node 之间路径上 的 所有 的 节点 存储到stack中
     *
     * @param root
     * @param node
     * @param stack
     * @return
     */
    private boolean getPath(TreeNode root, TreeNode node, Stack<TreeNode> stack) {
        if(root == null || node == null) {
            return false;
        }
        stack.push(root);
        if(root == node) {
            return true;
        }
        boolean flg = getPath(root.left,node,stack);
        if(flg) {
            return true;
        }
        boolean flg2 = getPath(root.right,node,stack);
        if(flg2) {
            return true;
        }
        stack.pop();
        return false;
    }

    public TreeNode lowestCommonAncestor2(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null) {
            return null;
        }
        Stack<TreeNode> stackP = new Stack<>();
        Stack<TreeNode> stackQ = new Stack<>();

        getPath(root,p,stackP);
        getPath(root,q,stackQ);

        //对栈的操作
        int sizeP = stackP.size();
        int sizeQ = stackQ.size();

        if(sizeP > sizeQ) {
            int size = sizeP - sizeQ;
            while (size != 0) {
                stackP.pop();
                size--;
            }
        }else {
            int size = sizeQ - sizeP;
            while (size != 0) {
                stackQ.pop();
                size--;
            }
        }
        //两个栈当中 元素的个数是相同的
        while (!stackP.isEmpty() && !stackQ.isEmpty()) {
            if(stackP.peek() .equals(stackQ.peek())) {
                return stackP.peek();
            }
            stackP.pop();
            stackQ.pop();
        }
        return null;
    }

12. 根据一棵树的前序遍历与中序遍历构造二叉树。

. - 力扣（LeetCode）

先序遍历确定根，根据中序遍历确定左树和右树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {


        public int priIndex;

        public TreeNode buildTree(int[] preorder, int[] inorder) {

            return buildTreeChild(preorder,inorder,0,inorder.length-1);
        }

        private TreeNode buildTreeChild(int[] preorder,int[] inorder,int inbegin,int inend) {

            //1. 没有左树 或者 没有右树了
            if(inbegin > inend) {
                return null;
            }
            //2.创建根节点
            TreeNode root = new TreeNode(preorder[priIndex]);

            //3.从中序遍历当中 找到根节点所在的下标
            int rootIndex = findIndex(inorder,inbegin,inend,preorder[priIndex]);
            if(rootIndex == -1) {
                return null;
            }

            priIndex++;
            //4. 创建左子树 和  右子树
            root.left = buildTreeChild(preorder,inorder,inbegin,rootIndex-1);

            root.right = buildTreeChild(preorder,inorder,rootIndex+1,inend);

            return root;
        }

        private int findIndex(int[] inorder,int inbegin,int inend,int key) {
            for(int i = inbegin;i <= inend;i++) {
                if(inorder[i] == key) {
                    return i;
                }
            }
            return -1;
        }
    }

13.根据一棵树的中序遍历与后序遍历构造二叉树

. - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
 class Solution {


    public int postIndex ;
    public TreeNode buildTree(int[] inorder, int[] postorder) {

        postIndex = postorder.length-1;

        return buildTreeChild(postorder,inorder,0,inorder.length-1);
    }

    private TreeNode buildTreeChild(int[] postorder,int[] inorder,int inbegin,int inend) {

        //1. 没有左树 或者 没有右树了
        if(inbegin > inend) {
            return null;
        }
        //2.创建根节点
        TreeNode root = new TreeNode(postorder[postIndex]);

        //3.从中序遍历当中 找到根节点所在的下标
        int rootIndex = findIndex(inorder,inbegin,inend,postorder[postIndex]);
        if(rootIndex == -1) {
            return null;
        }

        postIndex--;
        //4. 创建左子树 和  右子树

        root.right = buildTreeChild(postorder,inorder,rootIndex+1,inend);

        root.left = buildTreeChild(postorder,inorder,inbegin,rootIndex-1);

        return root;
    }

    private int findIndex(int[] inorder,int inbegin,int inend,int key) {
        for(int i = inbegin;i <= inend;i++) {
            if(inorder[i] == key) {
                return i;
            }
        }
        return -1;
    }
}

14.二叉树创建字符串。

. - 力扣（LeetCode）

public String tree2str(TreeNode root) {
        StringBuilder stringBuilder  = new StringBuilder();
        tree2strChild(root,stringBuilder);
        return stringBuilder.toString();
    }

    private void tree2strChild(TreeNode t,StringBuilder stringBuilder) {
        if(t == null) {
            return;
        }
        stringBuilder.append(t.val);
        if(t.left != null) {
            stringBuilder.append("(");
            tree2strChild(t.left,stringBuilder);
            stringBuilder.append(")");
        }else {
            if(t.right == null) {
                return;
            }else{
                stringBuilder.append("()");
            }
        }
        if(t.right != null) {
            stringBuilder.append("(");
            tree2strChild(t.right,stringBuilder);
            stringBuilder.append(")");
        }else {
            return;
        }
    }

    public void preOrderNor(TreeNode root) {
        if(root == null) return;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode top = null;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                System.out.print(cur.val + " ");
                cur = cur.left;
            }

            top = stack.pop();
            cur = top.right;
        }
    }

    public void inOrderNor(TreeNode root) {
        if(root == null) return;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode top = null;
        while (cur != null || !stack.isEmpty()) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            top = stack.pop();
            System.out.print(top.val + " ");
            cur = top.right;
        }
    }