LeetCode: Symmetric Tree 解题报告

Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   /
  2   2
 / /
3  4 4  3
But the following is not:
    1
   /
  2   2
     
   3    3
Note:
Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

SOL 1 & 2:

两个解法，递归和非递归.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    // solution 1:
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        
        return isSymmetricTree(root.left, root.right);
    }
    
    /*
     * 判断两个树是否互相镜像 
        (1) 根必须同时为空，或是同时不为空
     * 
     * 如果根不为空: 
        (1).根的值一样 
        (2).r1的左树是r2的右树的镜像 
        (3).r1的右树是r2的左树的镜像
     */
    public boolean isSymmetricTree1(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return true;
        }
        
        if (root1 == null || root2 == null) {
            return false;
        }
        
        return root1.val == root2.val && 
             isSymmetricTree(root1.left, root2.right) && 
             isSymmetricTree(root1.right, root2.left);
    }
    
    // solution 2:
    public boolean isSymmetricTree(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return true;
        }
        
        if (root1 == null || root2 == null) {
            return false;
        }
        
        Stack<TreeNode> s1 = new Stack<TreeNode>();
        Stack<TreeNode> s2 = new Stack<TreeNode>();
        
        // 一定记得初始化
        s1.push(root1);
        s2.push(root2);
        
        while (!s1.isEmpty() && !s2.isEmpty()) {
            TreeNode cur1 = s1.pop();
            TreeNode cur2 = s2.pop();
            
            if (cur1.val != cur2.val) {
                return false;
            }
            
            if (cur1.left != null && cur2.right != null) {
                s1.push(cur1.left);
                s2.push(cur2.right);
            } else if (!(cur1.left == null && cur2.right == null)) {
                return false;
            }
            
            if (cur1.right != null && cur2.left != null) {
                s1.push(cur1.right);
                s2.push(cur2.left);
            } else if (!(cur1.right == null && cur2.left == null)) {
                return false;
            }
        }
        
        return true;
    }
}

2015.1.20:

简化了递归的解法，root与root本身是一对对称树。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        return isMirror(root, root);
    }
    
    public boolean isMirror(TreeNode r1, TreeNode r2) {
        if (r1 == null && r2 == null) {
            return true;
        } else if (r1 == null || r2 == null) {
            return false;
        }
        
        return r1.val == r2.val
             && isMirror(r1.left, r2.right)
             && isMirror(r1.right, r2.left);
    }
}

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/IsSymmetric.java