<Tree> 110 124

110. Balanced Binary Tree

方法是如果我们发现子树不平衡，则不计算具体的深度，而是直接返回-1。那么优化后的方法为：对于每一个节点，我们通过checkDepth方法递归获得左右子树的深度，如果子树是平衡的，则返回真实的深度，若不平衡，直接返回-1，此方法时间复杂度O(N)，空间复杂度O(n)

class Solution {
    public boolean isBalanced(TreeNode root) {
        return height(root) != -1;
    }

    private int height(TreeNode root){
        if(root == null)
            return 0;
        int leftHeight = height(root.left);
        if(leftHeight == -1)
            return -1;
        int rightHeight = height(root.right);
        if(rightHeight == -1)
            return -1;
        if(leftHeight - rightHeight < -1 || leftHeight - rightHeight > 1)
            return -1;
        return Math.max(leftHeight, rightHeight) + 1;
    }
}

124. Binary Tree Maximum Path Sum

只有根节点才能包括左右子树的数值，如果是return则只能保存到左右子树中最大的一个分支

class Solution {

    public int maxPathSum(TreeNode root) {
        if(root == null) return 0;
        int[] maxPath =new int[] {Integer.MIN_VALUE};
        dfs(root, maxPath);
        return maxPath[0];
    }

    private int dfs(TreeNode root, int[] maxPath){
        if(root == null) return 0;
        int left = Math.max(dfs(root.left, maxPath), 0);
        int right = Math.max(dfs(root.right, maxPath), 0);
        maxPath[0] = Math.max(maxPath[0], root.val + left + right);
        return Math.max(left, right) + root.val;
    }
}