二叉树04

110.平衡二叉树 

题目链接：110. 平衡二叉树 - 力扣（LeetCode）

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

    private int getHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = getHeight(root.left);
        if (leftHeight == -1) {
            return -1;
        }
        int rightHeight = getHeight(root.right);
        if (rightHeight == -1) {
            return -1;
        }
        // 左右子树高度差大于1，return -1表示已经不是平衡树了
        if (Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        }
        return Math.max(leftHeight, rightHeight) + 1;
    }
}

 257. 二叉树的所有路径

题目链接：257. 二叉树的所有路径 - 力扣（LeetCode）

思路：回溯，就是递归回来后的处理

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> res = new ArrayList<>();
        if(root  == null){
            return res;
        }

        List<Integer> paths = new ArrayList<>();
        traversal(root,paths,res);
        return res;
    }

    private void traversal (TreeNode root, List<Integer> paths,List<String>res){
        //前序遍历
        paths.add(root.val);
        if(root.left == null && root.right == null){
            //输出
            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));
            res.add(sb.toString());
            return;
        }

        if(root.left != null){
            traversal(root.left,paths,res);
            paths.remove(paths.size() -1);
        }

        if(root.right != null){
            traversal(root.right,paths,res);
            paths.remove(paths.size() -1);
        }
    }
}

404.左叶子之和

题目链接：. - 力扣（LeetCode）

思路：关键在于判断出是左叶子节点，在父节点时判断

class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if(root == null) return 0;
        int leftValue = sumOfLeftLeaves(root.left);
        int rightValue = sumOfLeftLeaves(root.right);

        int midValue = 0;
        if(root.left != null && root.left.left == null && root.left.right == null)
            midValue = root.left.val;
        return midValue + leftValue + rightValue;
        
    }
}