剑指Offer之Java算法习题精讲二叉树专项训练

题目一

 解法

/**
 * 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 {
    int i = 0;
    int res = 0;
    public int kthSmallest(TreeNode root, int k) {
        method(root,k);
        return res;
    }
     public void method(TreeNode root, int k){
        if(root==null) return ;
        method(root.left,k);
        i++;
        if(i==k){
            res = root.val;
            return ;
        }
        method(root.right,k);
     }
}

题目二

 解法

/**
 * 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 {
    int sum = 0;
    public TreeNode convertBST(TreeNode root) {
        method(root);
        return root;
    }
    public void method(TreeNode root) {
        if(root==null){
            return;
        }
        method(root.right);
        sum+=root.val;
        root.val = sum;
        method(root.left);
 
    }
}

题目三

 解法

/**
 * 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 isValidBST(TreeNode root) {
        return method(root,null,null);
    }
    public boolean method(TreeNode root,TreeNode min,TreeNode max){
        if(root==null) return true;
        if(min!=null&&root.val<=min.val) return false;
        if(max!=null&&root.val>=max.val) return false;
        return method(root.left,min,root)&&method(root.right,root,max);
    }
}

题目四

解法

/**
 * 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 searchBST(TreeNode root, int val) {
        if(root==null) return null;
        if(root.val==val) return root;
        if(root.val>=val){
            return searchBST(root.left,val);
        }
        if(root.val 
 

题目五

 解法

/**
 * 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 insertIntoBST(TreeNode root, int val) {
        return  method(root,val);
    }
    public TreeNode method(TreeNode root, int val){
        if(root==null) return new TreeNode(val);
        if (root.val < val) 
            root.right = insertIntoBST(root.right, val);
        if (root.val > val) 
            root.left = insertIntoBST(root.left, val);
        return root;
    }
}

题目六

算法

/**
 * 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 deleteNode(TreeNode root, int key) {
        if (root == null) return null;
        if (root.val == key){
            if (root.left == null) return root.right;
            if (root.right == null) return root.left;
            TreeNode minNode = getMin(root.right);
            root.right = deleteNode(root.right, minNode.val);
            minNode.left = root.left;
            minNode.right = root.right;
            root = minNode;
        }else if(root.val>key){
            root.left = deleteNode(root.left,key);
        }else{
            root.right = deleteNode(root.right,key);
        }
        return root;
    }
    TreeNode getMin(TreeNode node) {
        while (node.left != null) node = node.left;
        return node;
    }   
}

到此这篇关于剑指Offer之Java算法习题精讲二叉树专项训练的文章就介绍到这了,更多相关Java 二叉树内容请搜索脚本之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持脚本之家！