樹的子結構
知識點:二叉樹
題目:
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)
解析:
- 思路:
代碼:
/**
public 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) {
//result作為最後的結果傳回
boolean result = false;
if (root1 != null && root2 != null) {
if (root1.val == root2.val) {
//判斷root1是否包含root2
result = doesTree1HaveTree2(root1, root2);
}
//root的左子樹上是否符合
if (!result) {
result = HasSubtree(root1.left, root2);
}
//root的右子樹上是否符合
if (!result) {
result = HasSubtree(root1.right, root2);
}
}
return result;
}
public boolean doesTree1HaveTree2(TreeNode node1, TreeNode node2) {
//如果node2周遊完都能對應上,傳回true
//這個判斷需要放在下面一個判斷的前面
if (node2 == null) {
return true;
}
if (node1 == null) {
return false;
}
//如果node1.val!=node2.val, 傳回false
if (node1.val != node2.val)
return false;
//如果根結點對應上,則分别在子結點上去比對
return doesTree1HaveTree2(node1.right, node2.right) && doesTree1HaveTree2(node1.left, node2.left);
}
}