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  • 18:树的子结构

    /**
     * 面试题18:树的子结构
     * 输入两棵二叉树A,B,判断B是不是A的子结构。(ps:我们约定空树不是任意一个树的子结构)
     */
    public class _18_tree_structure {
        public static void main(String[] args) {
            TreeNode18 t1 = new TreeNode18(1);
            t1.left=new TreeNode18(2);
            t1.right=new TreeNode18(3);
            t1.left.left=new TreeNode18(4);
            t1.left.right=new TreeNode18(5);
            TreeNode18 t2 = new TreeNode18(2);
            t2.left=new TreeNode18(4);
            t2.right=new TreeNode18(65);
    
            Solution18 solution18 = new Solution18();
            System.out.println(solution18.HasSubtree(t1,t2));
        }
    }
    
    class TreeNode18 {
        int val = 0;
        TreeNode18 left = null;
        TreeNode18 right = null;
    
        public TreeNode18(int val) {
            this.val = val;
        }
    }
    
    class Solution18 {
        public boolean HasSubtree(TreeNode18 root1, TreeNode18 root2) {
            boolean result = false;
            if(root1!=null && root2!=null){
                if(root1.val==root2.val){
                    result=recursionTree(root1,root2);
                }
                if(!result){
                    result=recursionTree(root1.left,root2);
                }
                if(!result){
                    result=recursionTree(root1.right,root2);
                }
            }
            return result;
        }
        public boolean recursionTree(TreeNode18 root1, TreeNode18 root2){
            if(root2==null){
                return true;
            }
            if(root1==null){
                return false;
            }
            if(root1.val!=root2.val){
                return false;
            }
            return recursionTree(root1.left,root2.left)&&recursionTree(root1.right,root2.right);
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/andy-zhou/p/6549298.html
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