Poison

110. Balanced Binary Tree

发表于
DFS
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class Solution {
    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        }

        if (Math.abs(getHeight(root.left) - getHeight(root.right)) > 1) {
            return false;
        }

        return isBalanced(root.left) && isBalanced(root.right);
    }

    private int getHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }

        return Math.max(getHeight(root.left), getHeight(root.right)) + 1;
    }
}

该方法最为直观，但是效率较低。

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

    /**
     * 返回当前二叉树的高度，如果当前二叉树不平衡，则返回 -1
     */
    private int getHeight(TreeNode node) {
        if (node == null) {
            return 0;
        }

        int leftHeight = getHeight(node.left);
        if (leftHeight == -1) {
            return -1;
        }

        int rightHeight = getHeight(node.right);
        if (rightHeight == -1) {
            return -1;
        }

        int diff = Math.abs(leftHeight - rightHeight);
        if (diff > 1) {
            return -1;
        } else {
            return Math.max(leftHeight, rightHeight) + 1; // 注意此处返回左右子树较大的高度并加 1, 即加上 root 节点
        }
    }
}

自底向上计算，如果遇到不平衡的子树则及时返回。

Reference

110. Balanced Binary Tree
剑指 Offer 55 - II. 平衡二叉树