【二叉树遍历和练习】

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一、二叉树前中后遍历

public class BinaryTree {
    static class TreeNode{
        public char val;
        public TreeNode left;
        public TreeNode right;

        public TreeNode(char val) {
            this.val = val;
        }

    }
    public TreeNode creatTree(){
        TreeNode A = new TreeNode('A');
        TreeNode B = new TreeNode('B');
        TreeNode C = new TreeNode('C');
        TreeNode D = new TreeNode('D');
        TreeNode E = new TreeNode('E');
        TreeNode F = new TreeNode('F');
        TreeNode G = new TreeNode('G');
        TreeNode H = new TreeNode('H');
        A.left = B;
        A.right = C;
        B.left = D;
        B.right = E;
        C.left = F;
        C.right = G;
        E.right = H;

        return A;
    }
    //前序遍历
    void preOrder(TreeNode root){
        //根左右
        if(root == null){
            return;
        }
        System.out.print(root.val+" ");
        preOrder(root.left);
        preOrder(root.right);

    }

    //中序遍历
    void inOrder(TreeNode root){
        //左根右
        if(root == null){
            return;
        }
        inOrder(root.left);
        System.out.print(root.val+" ");
        inOrder(root.right);

    }
    //后序遍历
    void postOrder(TreeNode root){
        //左右根
        if(root == null){
            return;
        }
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val+" ");
    }
    
        public static void main(String[] args) {
            BinaryTree binaryTree = new BinaryTree();
            BinaryTree.TreeNode root = binaryTree.creatTree();
            binaryTree.preOrder(root);
            System.out.println();
            binaryTree.inOrder(root);
            System.out.println();
            binaryTree.postOrder(root);
    }
}

二、获取节点个数

    public int nodeSize;
    int size(TreeNode root){
        if (root == null){
            return 0;
        }
        nodeSize++;
        size(root.left);
        size(root.right);
        return nodeSize;
    }
    //子问题思路解决：整棵树的节点=左子树的节点+右子数的节点+1
    int size2(TreeNode root){
        if (root == null){
            return 0;
        }
        return size2(root.left)+size2(root.right)+1;
    }

三.获取叶子节点个数

public int leafSize;
    int getLeafNodeCount(TreeNode root){
        if (root == null){
            return 0;
        }
        if (root.left == null && root.right == null){
            leafSize++;
        }
        getLeafNodeCount(root.left);
        getLeafNodeCount(root.right);
        return leafSize;
    }

四.获取第k层节点个数

 public int leveSize;
    int getleveNodeCount(TreeNode root,int k){
        if(root == null){
            return 0;
        }
        if (k == 1) {
            return 1;
        }
        return getleveNodeCount(root.left,k-1)
                + getleveNodeCount(root.right,k-1);
    }

五.求二叉树的高度，时间复杂度O（N）

int getHeight(TreeNode root){
        if (root == null){
            return 0;
        }
        int leftHeight = getHeight(root.left);
        int rightHeight = getHeight(root.right);
        return leftHeight > rightHeight?leftHeight+1:rightHeight+1;
    }

六.检测值为value的元素是否存在

TreeNode find(TreeNode root,int val){
        if (root == null){
            return null;
        }
        if (root.val == val){
            return root;
        }
        TreeNode ret1 = find(root.left,val);
        if (ret1 != null){
            return ret1;
        }
        TreeNode ret2 = find(root.right,val);
        if (ret2 != null){
            return ret2;
        }
        return null;
    }

七. 检查两颗树是否相同

    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);
    }

八.判断一棵二叉树是不是平衡二叉树

时间复杂度O（N）

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

九.一个二叉树的根节点 root ， 检查它是否轴对称

public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return isSymmetricChild(root.left,root.right);
    }
        private boolean isSymmetricChild(TreeNode leftTree,TreeNode rightTree){
            if((leftTree == null && rightTree != null) ||(leftTree != null && rightTree == null)){
                return false;
            }
            if(leftTree == null && rightTree == null){
                return true;
            }
            if(leftTree.val != rightTree.val){
                return false;
            }
            return isSymmetricChild(leftTree.left,rightTree.right) && isSymmetricChild(leftTree.right,rightTree.left);

        }

十. 判断subRoot是不是root的子树

    //时间复杂度：O(m*n)
    public boolean isSubtree(TreeNode root,TreeNode subTree) {
        if(root == null || subTree == null) {
            return false;
        }
        //判断根节点是否相同
        if (isSameTree(root,subTree)){
            return true;
        }
        //判断左子树是否相同
        if(isSubtree(root.left, subTree)) {
            return true;
        }
        //判断右子树是否相同
        if(isSubtree(root.right, subTree)) {
            return true;
        }
        return false;

    }

        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);
        }

十一.翻转二叉树

    public TreeNode invertTree(TreeNode root){
        if(root == null){
            return null;
        }
        if(root.left == null && root.right == null){
            return root;
        }
        //交换左右节点
        TreeNode tmp = root.left;
        root.left = root.right;
        root.right = tmp;
        //交换左树
        invertTree(root.left);
        //交换右树
        invertTree(root.right);

        return root;
    }

总结

今天复习二叉树练习。
时隔一个月继续更新文章，学校的考试周太可怕了=.= 怕挂科所以全身心投入复习，平时都学编程了，学校的课靠最后几周学完全部，也是挺厉害的哈哈。