题目来源

101. 对称二叉树 - 力扣（LeetCode）

代码1(递归)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return isMirror(root.left,root.right);
    }
    //判断两是否对称
    private boolean isMirror(TreeNode left, TreeNode right) {
        if(left==null && right==null) return true;
        //只有一个为空
        if(left==null || right==null) return false;
        if(left.val != right.val) return false;
        // 左左与右右 左右与右左
        return isMirror(left.left,right.right) && isMirror(left.right,right.left);
    }
}

代码分析

两子树是否对称最小规模子问题就是： 

1. 左子树 与 右子树是否堆成

（对称：左子树与右子树是否都为空，两值是否都为空， 不对称：只有一值为空）

2. 左子树的左子树与 右子树的右子树  是否同时与  左子树的右子树与 右子树的左子树 均为对称

代码2（dfs迭代法）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;

        Stack<TreeNode> stack = new Stack<>();
        stack.push(root.right);
        stack.push(root.left);

        while(!stack.isEmpty()) {
            TreeNode node1 = stack.pop(); //左子树根结点
            TreeNode node2 = stack.pop(); //右

            if(node1==null && node2==null) continue;
            if(node1==null || node2==null || node1.val!=node2.val) return false;
            // if(node1==null || node2==null) return false;
            // if( node1.val!=node2.val) return false;
            //将......逆向入栈 ： 左左与右右  左右与右左
            // 入栈就是为了 看两子树是否对称
            stack.push(node2.left);
            stack.push(node1.right);
            stack.push(node2.right);
            stack.push(node1.left);
        }
        return true;
    }
}

代码分析

这里的dfs，是双向深入的dfs。因为需要两子树进行比较。

代码3（bfs迭代法）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);

        while(!queue.isEmpty()) {
            TreeNode node1 = queue.poll();
            TreeNode node2 = queue.poll();

            if(node1==null && node2==null) continue;
            if(node1==null || node2==null || node1.val!=node2.val) return false;

            queue.offer(node1.left);
            queue.offer(node2.right);
            queue.offer(node1.right);
            queue.offer(node2.left);
        }
        return true;
    }
}

代码分析

类似dfs，只是逐层处理。