Symmetric Tree

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / \
  2   2
   \   \
   3    3

Note:

Bonus points if you could solve it both recursively and iteratively.

Tags

Tree, DFS, BFS

----------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Recursive Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function (root) {
    if (root === null) {
        return true;
    }

    return symmetric(root.left, root.right);
};

function symmetric(left, right) {
    if (left === null && right === null) {
        return true;
    }
    else if (left === null || right === null) {
        return false;
    } else {
        return (left.val === right.val) && symmetric(left.left, right.right) && symmetric(left.right, right.left);
    }
}

Iterative Solution

Using BFS to check for each level,

if the level of the queue is not multiple of two, return false;

other wise using two pointers to check at each end of the array, if the value is not equal, return false;

*** Pay attention to null left/right node, they also needs to push into the queue. 

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    if(root === null || (root.left === null && root.right === null)) {
        return true;
    }
    
    if(root.left === null || root.right === null) {
        return false;
    }
    
    // Guarantee that root has both left and right
    var queue = [];
    queue.push(root.left);
    queue.push(root.right);
    
    while(queue.length > 0) {
        var len = queue.length;
        if(len %2 !== 0) {
            return false;
        }
        
        var start = 0, end = len-1;
        while(start < end) {
            if(queue[start] === null && queue[end] === null) {
                start++;
                end--;
            } else if(queue[start] === null || queue[end] === null) {
                return false;
            } else if(queue[start].val !== queue[end].val) {
                return false;
            } else {
                start++;
                end--;
            }
        }
        
        var currentLevel = [];
        for(var i in queue) {
            if(queue[i] !== null) {
                currentLevel.push(queue[i].left);
                currentLevel.push(queue[i].right);
            }
        }
        queue = currentLevel;
    } 
    
    return true;
};