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404. Sum of Left Leaves

时间:2017-10-22 01:40:25      阅读:164      评论:0      收藏:0      [点我收藏+]

标签:while   logs   empty   public   stack   ack   ret   tin   col   

Find the sum of all left leaves in a given binary tree.

Example:

    3
   /   9  20
    /     15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
题目含义:获取所有左叶子节点的总和
方法一:
 1     public int sumOfLeftLeaves(TreeNode root) {
 2         if (root == null) return 0;
 3         int ans = 0;
 4         if (root.left != null) {
 5             if (root.left.left == null && root.left.right == null) ans += root.left.val;
 6             else ans += sumOfLeftLeaves(root.left);
 7         }
 8         ans += sumOfLeftLeaves(root.right);
 9         return ans;
10     }

方法二:

 1     public int sumOfLeftLeaves(TreeNode root) {
 2         if (root == null) return 0;
 3         int ans = 0;
 4         Stack<TreeNode> stack = new Stack<TreeNode>();
 5         stack.push(root);
 6         while (!stack.empty()) {
 7             TreeNode node = stack.pop();
 8             if (node.left != null) {
 9                 if (node.left.left == null && node.left.right == null) ans += node.left.val;
10                 else stack.push(node.left);
11             }
12             if (node.right != null) {
13                 if (node.right.left == null && node.right.right == null) continue;
14                 else stack.push(node.right);
15             }
16         }
17         return ans;
18     }

 

404. Sum of Left Leaves

标签:while   logs   empty   public   stack   ack   ret   tin   col   

原文地址:http://www.cnblogs.com/wzj4858/p/7707340.html

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