剑指Offer——题26（树的子结构）

最新推荐文章于 2022-07-05 17:23:03 发布

原创最新推荐文章于 2022-07-05 17:23:03 发布 · 82 阅读

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本文介绍了一种算法，用于判断一棵二叉树B是否为另一棵二叉树A的子结构。通过递归方式检查A树中是否存在与B树根节点相同的节点，并进一步判断以此节点为根的子树是否与B树完全相同。文章提供了详细的Java代码实现及测试案例。

1.题目

输入两棵二叉树A和B，判断B是不是A的子结构

2 .思路

step1：判断A树中是否含有B树的根节点R

step2：判断对应根节点下两棵子树是否相等

注意：判断浮点数相等不能直接用 “==“ 判断。

3.代码实现

public class HasSubtree {

    class TreeNode {
        int val = 0;
        TreeNode left = null;
        TreeNode right = null;

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

        public TreeNode(int val, TreeNode left, TreeNode right) {
            this.val = val;
            this.left = left;
            this.right = right;
        }
    }


    public boolean HasSubtree(TreeNode root1,TreeNode root2) {
        boolean isEquals=false;
        if(root1!=null&&root2!=null){

            if(equals(root1,root2)){
                isEquals=doesTree1HasTree2(root1, root2);
            }
            if(!isEquals){
                isEquals=HasSubtree(root1.left,root2);
            }
            if(!isEquals){
                isEquals=HasSubtree(root1.right,root2);
            }
        }
        return isEquals;
    }

    private boolean doesTree1HasTree2(TreeNode root1, TreeNode root2) {
        if(root2==null){
            return true;
        }
        if(root1==null){
            return false;
        }
        if(!equals(root1,root2)){
            return false;
        }
        return doesTree1HasTree2(root1.right,root2.right)&&doesTree1HasTree2(root1.left,root2.left);
    }


    public boolean equals(TreeNode root1,TreeNode root2){
        if(root1.val-root2.val<0.0000001&&root1.val-root2.val>-0.0000001){
            return true;
        }
        return false;
    }

    //正常功能测试
    @Test
    public void test1() {
        TreeNode treeNode1 = new TreeNode(8);
        TreeNode treeNode2 = new TreeNode(8);
        TreeNode treeNode3 = new TreeNode(7);
        TreeNode treeNode4 = new TreeNode(9);
        TreeNode treeNode5 = new TreeNode(2);
        TreeNode treeNode6 = new TreeNode(4);
        TreeNode treeNode7 = new TreeNode(7);

        treeNode1.left=treeNode2;
        treeNode1.right=treeNode3;
        treeNode2.left=treeNode4;
        treeNode2.right=treeNode5;
        treeNode5.left=treeNode6;
        treeNode5.right=treeNode7;


        TreeNode tree2Node1 = new TreeNode(8);
        TreeNode tree2Node2 = new TreeNode(9);
        TreeNode tree2Node3 = new TreeNode(2);
        tree2Node1.left=tree2Node2;
        tree2Node1.right=tree2Node3;
        boolean isExist=HasSubtree(treeNode1,tree2Node1);
        System.out.println(isExist);

    }

//A树只有一个节点
    @Test
    public void test2() {
        TreeNode treeNode1 = new TreeNode(8);

        TreeNode tree2Node1 = new TreeNode(8);
        TreeNode tree2Node2 = new TreeNode(9);
        TreeNode tree2Node3 = new TreeNode(2);
        tree2Node1.left=tree2Node2;
        tree2Node1.right=tree2Node3;
        boolean isExist=HasSubtree(treeNode1,tree2Node1);
        System.out.println(isExist);
    }

    //A树为空
    @Test
    public void test3() {
        TreeNode tree2Node1 = new TreeNode(8);
        TreeNode tree2Node2 = new TreeNode(9);
        TreeNode tree2Node3 = new TreeNode(2);
        tree2Node1.left=tree2Node2;
        tree2Node1.right=tree2Node3;
        boolean isExist=HasSubtree(null,tree2Node1);
        System.out.println(isExist);
    }
}