[剑指offer]判断两个二叉树是否相等及变形

剑指offer 26，27，28 这三道题代码结构相似，可以放在一起练习。

 

剑指offer 28  对称的二叉树

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return dfs(root.left, root.right);
    }
    public boolean dfs(TreeNode p, TreeNode q){
        if(p == null && q == null) return true;
        if(p == null || q == null) return false;
        if(p.val != q.val) return false;
        return dfs(p.left, q.right) && dfs(p.right, q.left);
    }
}

 

剑指offer 27  二叉树的镜像

class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root == null) return null;
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        mirrorTree(root.left);
        mirrorTree(root.right);
        return root;
    }
}

 

剑指offer 26  树的子结构

class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if(A == null || B == null) return false;
        if(dfs(A, B)){
            return true;
        }
        return isSubStructure(A.left, B) || isSubStructure(A.right, B);
    }
    public boolean dfs(TreeNode p, TreeNode q){
        if(q == null) return true;
        if(p == null || p.val != q.val) return false;
        return dfs(p.left, q.left) && dfs(p.right, q.right);
    }
}