404. 左叶子之和

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public int sumOfLeftLeaves1(TreeNode root) {
        if(root==null) return 0;
        //递归，左右中
        int leftVal = sumOfLeftLeaves(root.left);
        int rightVal = sumOfLeftLeaves(root.right);
        int medialVal = 0;
        //判断是否叶子节点
        //如果该节点的左节点不为空，该节点的左节点的左节点为空，该节点的左节点的右节点为空，则找到了一个左叶子，
        if(root.left!=null&&root.left.left==null&&root.left.right==null){
            medialVal += root.left.val;
        }
        return medialVal+leftVal+rightVal;
    }

    //本题迭代法使用前中后序都是可以的，只要把左叶子节点统计出来，就可以了
     public int sumOfLeftLeaves(TreeNode root) {
        if(root==null) return 0;
        //使用栈
        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 ;

     }
}
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