今天看啥
热点:

【剑指Offer学习】【面试题39:二叉树的深度】,剑指offer


题目一:输入一棵二叉树的根结点,求该树的深度。从根结点到叶子点依次经过的结点(含根、叶结点)形成树的一条路径,最长路径的长度为树的深度。


二叉树的结点定义

private static class BinaryTreeNode {
    int val;
    BinaryTreeNode left;
    BinaryTreeNode right;

    public BinaryTreeNode() {
    }

    public BinaryTreeNode(int val) {
        this.val = val;
    }
}

解题思路

  如果一棵树只有一个结点,它的深度为。 如果根结点只有左子树而没有右子树, 那么树的深度应该是其左子树的深度加1,同样如果根结点只有右子树而没有左子树,那么树的深度应该是其右子树的深度加1. 如果既有右子树又有左子树, 那该树的深度就是其左、右子树深度的较大值再加1 . 比如在图6.1 的二叉树中,根结点为1 的树有左右两个子树,其左右子树的根结点分别为结点2和3。根结点为2 的左子树的深度为3 , 而根结点为3 的右子树的深度为2,因此根结点为1的树的深度就是4 。
  这个思路用递归的方法很容易实现, 只儒对遍历的代码稍作修改即可。

这里写图片描述

代码实现

public static int treeDepth(BinaryTreeNode root) {
    if (root == null) {
        return 0;
    }

    int left = treeDepth(root.left);
    int right = treeDepth(root.right);

    return left > right ? (left + 1) : (right + 1);
}

题目二:输入一棵二叉树的根结点,判断该树是不是平衡二叉树。如果某二叉树中任意结点的左右子树的深度相差不超过1 ,那么它就是一棵平衡二叉树。


解题思路

解法一:需要重蟹遍历结点多次的解法
  在遍历树的每个结点的时候,调用函数treeDepth得到它的左右子树的深度。如果每个结点的左右子树的深度相差都不超过1 ,按照定义它就是一棵平衡的二叉树。

public static boolean isBalanced(BinaryTreeNode root) {
    if (root == null) {
        return true;
    }

    int left = treeDepth(root.left);
    int right = treeDepth(root.right);
    int diff = left - right;
    if (diff > 1 || diff < -1) {
        return false;
    }

    return isBalanced(root.left) && isBalanced(root.right);
}

解法二:每个结点只遍历一次的解法
  用后序遍历的方式遍历二叉树的每一个结点,在遍历到一个结点之前我们就已经遍历了它的左右子树。只要在遍历每个结点的时候记录它的深度(某一结点的深度等于它到叶节点的路径的长度),我们就可以一边遍历一边判断每个结点是不是平衡的。

/**
 * 判断是否是平衡二叉树,第二种解法
 *
 * @param root
 * @return
 */
public static boolean isBalanced2(BinaryTreeNode root) {
    int[] depth = new int[1];
    return isBalancedHelper(root, depth);
}

public static boolean isBalancedHelper(BinaryTreeNode root, int[] depth) {
    if (root == null) {
        depth[0] = 0;
        return true;
    }

    int[] left = new int[1];
    int[] right = new int[1];

    if (isBalancedHelper(root.left, left) && isBalancedHelper(root.right, right)) {
        int diff = left[0] - right[0];
        if (diff >= -1 && diff <= 1) {
            depth[0] = 1 + (left[0] > right[0] ? left[0] : right[0]);
            return true;
        }
    }

    return false;
}

完整代码

public class Test39 {

    private static class BinaryTreeNode {
        int val;
        BinaryTreeNode left;
        BinaryTreeNode right;

        public BinaryTreeNode() {
        }

        public BinaryTreeNode(int val) {
            this.val = val;
        }
    }

    public static int treeDepth(BinaryTreeNode root) {
        if (root == null) {
            return 0;
        }

        int left = treeDepth(root.left);
        int right = treeDepth(root.right);

        return left > right ? (left + 1) : (right + 1);
    }

    /**
     * 判断是否是平衡二叉树,第一种解法
     *
     * @param root
     * @return
     */
    public static boolean isBalanced(BinaryTreeNode root) {
        if (root == null) {
            return true;
        }

        int left = treeDepth(root.left);
        int right = treeDepth(root.right);
        int diff = left - right;
        if (diff > 1 || diff < -1) {
            return false;
        }

        return isBalanced(root.left) && isBalanced(root.right);
    }


    /**
     * 判断是否是平衡二叉树,第二种解法
     *
     * @param root
     * @return
     */
    public static boolean isBalanced2(BinaryTreeNode root) {
        int[] depth = new int[1];
        return isBalancedHelper(root, depth);
    }

    public static boolean isBalancedHelper(BinaryTreeNode root, int[] depth) {
        if (root == null) {
            depth[0] = 0;
            return true;
        }

        int[] left = new int[1];
        int[] right = new int[1];

        if (isBalancedHelper(root.left, left) && isBalancedHelper(root.right, right)) {
            int diff = left[0] - right[0];
            if (diff >= -1 && diff <= 1) {
                depth[0] = 1 + (left[0] > right[0] ? left[0] : right[0]);
                return true;
            }
        }

        return false;
    }

    public static void main(String[] args) {
        test1();
        test2();
        test3();
        test4();
    }


    // 完全二叉树
    //             1
    //         /      \
    //        2        3
    //       /\       / \
    //      4  5     6   7
    private static void test1() {
        BinaryTreeNode n1 = new BinaryTreeNode(1);
        BinaryTreeNode n2 = new BinaryTreeNode(1);
        BinaryTreeNode n3 = new BinaryTreeNode(1);
        BinaryTreeNode n4 = new BinaryTreeNode(1);
        BinaryTreeNode n5 = new BinaryTreeNode(1);
        BinaryTreeNode n6 = new BinaryTreeNode(1);
        BinaryTreeNode n7 = new BinaryTreeNode(1);

        n1.left = n2;
        n1.right = n3;
        n2.left = n4;
        n2.right = n5;
        n3.left = n6;
        n3.right = n7;

        System.out.println(isBalanced(n1));
        System.out.println(isBalanced2(n1));
        System.out.println("----------------");

    }

    // 不是完全二叉树,但是平衡二叉树
    //             1
    //         /      \
    //        2        3
    //       /\         \
    //      4  5         6
    //        /
    //       7
    private static void test2() {
        BinaryTreeNode n1 = new BinaryTreeNode(1);
        BinaryTreeNode n2 = new BinaryTreeNode(1);
        BinaryTreeNode n3 = new BinaryTreeNode(1);
        BinaryTreeNode n4 = new BinaryTreeNode(1);
        BinaryTreeNode n5 = new BinaryTreeNode(1);
        BinaryTreeNode n6 = new BinaryTreeNode(1);
        BinaryTreeNode n7 = new BinaryTreeNode(1);

        n1.left = n2;
        n1.right = n3;
        n2.left = n4;
        n2.right = n5;
        n5.left = n7;
        n3.right = n6;


        System.out.println(isBalanced(n1));
        System.out.println(isBalanced2(n1));
        System.out.println("----------------");
    }

    // 不是平衡二叉树
    //             1
    //         /      \
    //        2        3
    //       /\
    //      4  5
    //        /
    //       7
    private static void test3() {
        BinaryTreeNode n1 = new BinaryTreeNode(1);
        BinaryTreeNode n2 = new BinaryTreeNode(1);
        BinaryTreeNode n3 = new BinaryTreeNode(1);
        BinaryTreeNode n4 = new BinaryTreeNode(1);
        BinaryTreeNode n5 = new BinaryTreeNode(1);
        BinaryTreeNode n6 = new BinaryTreeNode(1);
        BinaryTreeNode n7 = new BinaryTreeNode(1);

        n1.left = n2;
        n1.right = n3;
        n2.left = n4;
        n2.right = n5;
        n5.left = n7;

        System.out.println(isBalanced(n1));
        System.out.println(isBalanced2(n1));
        System.out.println("----------------");
    }

    //               1
    //              /
    //             2
    //            /
    //           3
    //          /
    //         4
    //        /
    //       5
    private static void test4() {
        BinaryTreeNode n1 = new BinaryTreeNode(1);
        BinaryTreeNode n2 = new BinaryTreeNode(1);
        BinaryTreeNode n3 = new BinaryTreeNode(1);
        BinaryTreeNode n4 = new BinaryTreeNode(1);
        BinaryTreeNode n5 = new BinaryTreeNode(1);
        BinaryTreeNode n6 = new BinaryTreeNode(1);
        BinaryTreeNode n7 = new BinaryTreeNode(1);

        n1.left = n2;
        n2.left = n3;
        n3.left = n4;
        n4.left = n5;


        System.out.println(isBalanced(n1));
        System.out.println(isBalanced2(n1));
        System.out.println("----------------");
    }

    // 1
    //  \
    //   2
    //    \
    //     3
    //      \
    //       4
    //        \
    //         5
    private static void test5() {
        BinaryTreeNode n1 = new BinaryTreeNode(1);
        BinaryTreeNode n2 = new BinaryTreeNode(1);
        BinaryTreeNode n3 = new BinaryTreeNode(1);
        BinaryTreeNode n4 = new BinaryTreeNode(1);
        BinaryTreeNode n5 = new BinaryTreeNode(1);
        BinaryTreeNode n6 = new BinaryTreeNode(1);
        BinaryTreeNode n7 = new BinaryTreeNode(1);

        n1.right = n2;
        n2.right = n3;
        n3.right = n4;
        n4.right = n5;


        System.out.println(isBalanced(n1));
        System.out.println(isBalanced2(n1));
        System.out.println("----------------");
    }
}

运行结果

这里写图片描述

版权声明:本文为博主原创文章,未经博主允许不得转载。

www.bkjia.comtruehttp://www.bkjia.com/ASPjc/1027255.htmlTechArticle【剑指Offer学习】【面试题39:二叉树的深度】,剑指offer 题目一:输入一棵二叉树的根结点,求该树的深度。从根结点到叶子点依次经过的...

相关文章

帮客评论

视觉看点