leecode 404. 左叶子之和

404. 左叶子之和

        本题注意是左叶子节点，  不是左节点，因此需要判断是否是左叶子节点，当一个节点的左节点不为空，左节点的左右节点都为空时，则为左叶子节点。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */



迭代：

class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) return 0;
        Stack<TreeNode> stack = new Stack<> ();
        stack.add(root);
        int result = 0;
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (node.left != null && node.left.left == null && node.left.right == null) {
                result += node.left.val;
            }
            if (node.right != null) stack.add(node.right);
            if (node.left != null) stack.add(node.left);
        }
        return result;
    }
}


递归：

class Solution {
    public int sumOfLeftLeaves(TreeNode root) {
        if (root == null) return 0;
//l为左节点的左叶子节点数
//r为右节点的左叶子节点数
        int l = sumOfLeftLeaves(root.left);
        int r = sumOfLeftLeaves(root.right);
//当前节点的左叶子节点数,即如果当前节点左节点为叶子节点时，则m的值。
        int m = 0;
        if (root.left != null && root.left.left == null && root.left.right == null) m = root.left.val;
        int sum = l + r + m;
        return sum;
    }
}