• java实现二叉树常见操作


    package com.xk.test.struct.newp;
    
    import java.util.ArrayList;
    import java.util.LinkedList;
    import java.util.Queue;
    import java.util.Stack;
    
    public class MyBinaryTree {
    
        /**
         * 插入节点
         * @param root
         * @param node
         * @return
         */
        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;
        }
        
        /**
         * 递归解法前序遍历
         * @param root
         * @return
         */
        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);
        }
        
        /**
         * 迭代解法前序遍历
         * @param root
         * @return
         */
        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;
        }
            
        /**
         * 中序遍历
         * @param root
         * @return
         */
        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;
        }
        
        /**
         * 后序遍历
         * @param root
         * @return
         */
        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;
        }
        
        /**
         * 最大深度
         * @param node
         * @return
         */
        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;
        }
        
        /**
         * 层次遍历
         * @param root
         * @return
         */
        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;
        }
        
        /**
         * 最小深度
         * @param root
         * @return
         */
        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;
        }
        
        /**
         * 节点的个数
         * @param root
         * @return
         */
        int numOfTreeNode(TreeNode root){
            if(root == null){
                return 0;
                
            }
            int left = numOfTreeNode(root.left);
            int right = numOfTreeNode(root.right);
            return left + right + 1;
        }
        
        /**
         * 叶子节点的个数
         * @param root
         * @return
         */
        int numsOfNodeTreeNode(TreeNode root){
            if(root == null){
                return 0;
            }
            if(root.left==null&&root.right==null){
                return 1;
            }
            return numsOfNodeTreeNode(root.left)+numsOfNodeTreeNode(root.right);
            
        }
        
        /**
         * 第k层节点的个数
         * @param root
         * @param k
         * @return
         */
        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;
        }
        
        /**
         * 翻转二叉树or镜像二叉树
         * @param root
         * @return
         */
        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;
        }
        
        /**
         * 两个二叉树是否互为镜像
         * @param t1
         * @param t2
         * @return
         */
        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);
        }
        
        /**
         * 是否是平衡二叉树
         * 它是一棵空树或它的左右两个子树的高度差的绝对值不超过1,并且左右两个子树都是一棵平衡二叉树。
         * @param node
         * @return
         */
        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;
        }
        
        /**
         * 是否是完全二叉树
         * 对于深度为K的,有n个结点的二叉树,当且仅当其每一个结点都与深度为K的满二叉树中编号从1至n的结点一一对应时称之为完全二叉树。
         * @param root
         * @return
         */
        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;
        }
        
        
        /**
         * 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;
        }
        
        /**
         * 把二叉树打印成多行
         * @param pRoot
         * @return
         */
        ArrayList<ArrayList<Integer> > Print(TreeNode pRoot) {
            ArrayList<ArrayList<Integer> > res = new ArrayList<ArrayList<Integer> >();
            if(pRoot == null)
                return res;
            ArrayList<Integer> temp = new ArrayList<Integer>();
            Queue<TreeNode> layer = new LinkedList<TreeNode>();
            layer.offer(pRoot);
            int start = 0, end = 1;
            while(!layer.isEmpty()){
                TreeNode node = layer.poll();
                temp.add(node.val);
                start ++;
                if(node.left != null)
                    layer.add(node.left);
                if(node.right != null)
                    layer.add(node.right);
                if(start == end){
                    start = 0;
                    res.add(temp);
                    temp = new ArrayList<Integer>();
                    end = layer.size();
                }
            }
            return res;
        }
        
        /**
         * 按之字形顺序打印二叉树
         * @param pRoot
         * @return
         */
        public ArrayList<ArrayList<Integer> > PrintZ(TreeNode pRoot) {
            ArrayList<ArrayList<Integer> > res = new ArrayList<ArrayList<Integer> >();
            Stack<TreeNode> s1 = new Stack<TreeNode>();
            Stack<TreeNode> s2 = new Stack<TreeNode>();
            int flag = 1;
            if(pRoot == null)
                return res;
            s2.push(pRoot);
            ArrayList<Integer> temp = new ArrayList<Integer>();
            while(!s1.isEmpty() || !s2.isEmpty()){
                if(flag % 2 != 0){
                    while(!s2.isEmpty()){
                        TreeNode node = s2.pop();
                        temp.add(node.val);
                        if(node.left != null){
                            s1.push(node.left);
                        }
                        if(node.right != null){
                            s1.push(node.right);
                        }
                    }
                }
                if(flag % 2 == 0){
                    while(!s1.isEmpty()){
                        TreeNode node = s1.pop();
                        temp.add(node.val);
                        if(node.right != null){
                            s2.push(node.right);
                        }
                        if(node.left != null){
                            s2.push(node.left);
                        }
                    }
                }
                res.add(new ArrayList<Integer>(temp));
                temp.clear();
                flag ++;
            }
            return res;
        }
    
    }
    
    class TreeNode{
        int val;
        //左孩子
        TreeNode left;
        //右孩子
        TreeNode right;
    }

    转自:https://www.jianshu.com/p/0190985635eb

               https://www.weiweiblog.cn/printz/

    寻找撬动地球的支点(解决问题的方案),杠杆(Java等编程语言)已经有了。xkzhangsan
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  • 原文地址:https://www.cnblogs.com/xkzhangsanx/p/11001165.html
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