Same Tree

Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

难度：easy

Solution: DFS

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

        if(p == q || p == null || q == null|| p.val != q.val) {
            return false;
        }
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}

Symmetric Tree

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

难度：easy

Solution: DFS

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

Solution: BFS

BFS有很多种不同的用法，不同用法之间的差别一般是Queue的实现方式不一样。这道题用到了Deque，实际上是两个stack。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) {
            return true;
        }
        Deque<TreeNode> curLevel = new LinkedList<>();
        curLevel.add(root.left);
        curLevel.add(root.right);
        while(!curLevel.isEmpty()) {
            Deque<TreeNode> nextLevel = new LinkedList<>();
            while(!curLevel.isEmpty()) {
                TreeNode first = curLevel.removeFirst();
                TreeNode last = curLevel.removeLast();
                if(first == null && last == null) {
                    continue;
                }
                if(first == null || last == null || first.val != last.val) {
                    return false;
                }
                nextLevel.addFirst(first.left);
                nextLevel.addFirst(first.right);
                nextLevel.addLast(last.right);
                nextLevel.addLast(last.left);
            }
            curLevel = nextLevel;
        }
        return true;
    }
}