leetcode_Java101對稱二叉樹

1.遞歸

Their two roots have the same value.
The right subtree of each tree is a mirror reflection of the left subtree of the other tree.

Complexity Analysis

Time complexity : O(n)O(n). Because we traverse the entire input tree once, the total run time is O(n)O(n), where nn is the total number of nodes in the tree.

Space complexity : The number of recursive calls is bound by the height of the tree. In the worst case, the tree is linear and the height is in O(n)O(n). Therefore, space complexity due to recursive calls on the stack is O(n)O(n) in the worst case

public boolean isSymmetric(TreeNode root) {
    return isMirror(root, root);
}

public boolean isMirror(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) return true;
    if (t1 == null || t2 == null) return false;
    return (t1.val == t2.val)
        && isMirror(t1.right, t2.left)
        && isMirror(t1.left, t2.right);
}

解法2：迭代
Instead of recursion, we can also use iteration with the aid of a queue. Each two consecutive nodes in the queue should be equal, and their subtrees a mirror of each other. Initially, the queue contains root and root. Then the algorithm works similarly to BFS, with some key differences. Each time, two nodes are extracted and their values compared. Then, the right and left children of the two nodes are inserted in the queue in opposite order. The algorithm is done when either the queue is empty, or we detect that the tree is not symmetric (i.e. we pull out two consecutive nodes from the queue that are unequal).

 //利用一個隊列同時提取2個結點
    public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }
        Queue<TreeNode>  q = new LinkedList<>();
        q.add(root);
        q.add(root);
        while(!q.isEmpty()){
            TreeNode t1 = q.poll();
            TreeNode t2 = q.poll();
            if(t1 == null && t2 == null){
                continue;
            }
            if(t1 ==null || t2 ==null){
                return false;
            }
            
            if(t1.val != t2.val){
                return false;
            }
            
            q.add(t1.left);
            q.add(t2.right);
            q.add(t1.right);
            q.add(t2.left);
        }
        return true;
    }