重拾演算法之劍指Offier——樹的子結構
阿新 • • 發佈:2018-12-13
樹的子結構
題目描述
輸入兩棵二叉樹A,B,判斷B是不是A的子結構。(ps:我們約定空樹不是任意一個樹的子結構)
class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
public class Solution {
public boolean HasSubtree(TreeNode root1, TreeNode root2) {
boolean flag = false;
if (root1 != null && root2 != null){
if (root1.val == root2.val){
flag = doesTree1haveTree2(root1, root2);
}
if(flag == false){
flag = HasSubtree(root1.left, root2);
}
if(flag == false ){
flag = HasSubtree(root1.right, root2);
}
}
return flag;
}
public boolean doesTree1haveTree2(TreeNode root1, TreeNode root2){
if(root2 == null){
return true;
}
if(root1 == null || root1.val != root2.val){
return false;
}
return doesTree1haveTree2(root1.left,root2.left) && doesTree1haveTree2(root1.right,root2.right);
}
}