day17 打卡110.平衡二叉树 257. 二叉树的所有路径 404.左叶子之和

day17 打卡110.平衡二叉树 257. 二叉树的所有路径 404.左叶子之和

110.平衡二叉树

110题目链接

1.递归法

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

    public 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;
        }
        if (Math.abs(leftHeight-rightHeight) > 1) {
            return -1;
        }
        return Math.max(leftHeight, rightHeight)+1;
    }
}

257. 二叉树的所有路径

257题目链接

1.递归法，回溯

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<>();
        if (root == null) return result;
        // 作为结果中的路径
        List<Integer> paths = new ArrayList<>();
        traversal(root, paths, result);
        return result;
    }

    public void traversal(TreeNode node, List<Integer> paths, List<String> result) {
        paths.add(node.val);
        if (node.left == null && node.right == null) {
            StringBuilder sb = new StringBuilder();
            for (int i = 0 ; i<paths.size()-1 ; i++) {
                sb.append(paths.get(i)).append("->");
            }
            sb.append(paths.get(paths.size()-1));
            result.add(sb.toString());
            return;
        }
        // 递归和回溯是同时进行
        if (node.left != null) {
            traversal(node.left, paths, result);
            paths.remove(paths.size()-1);
        }
        if (node.right != null) {
            traversal(node.right, paths, result);
            paths.remove(paths.size()-1);
        }
    }
}

404.左叶子之和

404题目链接

1.递归法

class Solution {
    public int sumOfLeftLeaves(TreeNode node) {
        if (node == null) return 0;
        if (node.left == null && node.right == null) return 0;
        // 左
        int leftNum = sumOfLeftLeaves(node.left);
        // 右
        int rightNum = sumOfLeftLeaves(node.right);

        int midNum = 0;
        if (node.left != null && node.left.left == null && node.left.right == null) {
            midNum = node.left.val;
        }

        // 中
        return leftNum + rightNum + midNum;
    }
}

参考资料

代码随想录