Subject description:
Two binary inputs A, B, B is judged not substructure A's. (Ps: we agreed empty tree is not a tree any substructure)
Ideas: if A, B the same as the root value, which value is determined further subtree;
otherwise, the left subtree of A, find the right subtree
/* function TreeNode(x) {
this.val = x;
this.left = null;
this.right = null;
} */
function HasSubtree(pRoot1, pRoot2)
{
if (pRoot1 == null || pRoot2 == null) {
return false;
}
//如果根节点相同,进一步比较其左子树右子树;
//否则分别判断其左子树或右子树中是否包含
return Subtree(pRoot1, pRoot2) || HasSubtree(pRoot1.left, pRoot2) || HasSubtree(pRoot1.right, pRoot2);
}
function Subtree(root1, root2) {
//如果第二棵遍历结束,则说明是是其子树
if (root2 == null) return true;
//如果第二棵没有结束,第一棵已经遍历完,则不是它的子树
if (root1 == null) return false;
//如果两个节点值相同,进一步判断其左子树右子树
if (root1.val == root2.val) {
return Subtree(root1.left, root2.left) &&
Subtree(root1.right, root2.right);
} else {
return false;
}
}