Balanced Binary Tree

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Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

样例

Given binary tree A={3,9,20,#,#,15,7}, B={3,#,20,15,7}

A)  3            B)    3    / \                  \  9  20                 20    /  \                / \   15   7              15  7

The binary tree A is a height-balanced binary tree, but B is not.

思路：

看了网上的解法，好多的递归算法，时间复杂度应该为O(n),Depth first Search 空间上会比下面的算法节省，但是，要考虑JVM的栈溢出问题了。

总结下，递归算法，考虑问题简洁明了，但是也存在两个问题，一是递归栈溢出问题，二是，存在迭代计算问题，本例中为树，迭代为单项计算，所以不存在重复计算问题，但是如果是图，那么迭代的问题要慎重考虑了。

转换思路，a.将要递归的数据放到数据集合 b.使用map存储所有计算结果供以后计算用，时间复杂度O(2*n)

/** * Definition of TreeNode: * public class TreeNode { *     public int val; *     public TreeNode left, right; *     public TreeNode(int val) { *         this.val = val; *         this.left = this.right = null; *     } * } */public class Solution {    /**     * @param root: The root of binary tree.     * @return: True if this Binary tree is Balanced, or false.     */    public boolean isBalanced(TreeNode root) {if (root == null)return true;Queue<TreeNode> q = new LinkedList<TreeNode>();Stack<TreeNode> s = new Stack<TreeNode>();q.offer(root);s.push(root);while (!q.isEmpty()) {TreeNode f = q.poll();TreeNode l = f.left;TreeNode r = f.right;if (l != null) {q.offer(l);s.push(l);}if (r != null) {q.offer(r);s.push(r);}}Map<Integer, Integer> map = new HashMap<Integer, Integer>();while (!s.isEmpty()) {TreeNode c = s.pop();int l = 0;if(c.left!=null && map.get(c.left.val) != null)    l = map.get(c.left.val);int r = 0;if(c.right!=null && map.get(c.right.val) != null)    r = map.get(c.right.val);if ((l - r) * (l - r) > 1)return false;map.put(c.val, l > r ? l + 1 : r + 1);}return true;}}

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