LeetCode Sum of Left Leaves

原题链接在这里：https://leetcode.com/problems/sum-of-left-leaves/

题目：

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.

题解：

DFS, 若root.left不为null时，检查root.left是否为leaf. 若是res+=root.left.val, 若不是继续DFS.

再从root.right做DFS.

Time Complexity: O(n), n是tree的node数目. Space: O(logn).

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public int sumOfLeftLeaves(TreeNode root) {
         if(root == null){
             return 0;
         }

         int res = 0;
         if(root.left != null){
             if(root.left.left == null && root.left.right == null){
                 res += root.left.val;
             }else{
                 res += sumOfLeftLeaves(root.left);
             }
         }

         res += sumOfLeftLeaves(root.right);

         return res;
     }
 }

Iteration 做法.

Time Complexity: O(n). Space: O(logn).

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public int sumOfLeftLeaves(TreeNode root) {
         if(root == null){
             return 0;
         }

         int res = 0;
         Stack<TreeNode> stk = new Stack<TreeNode>();
         stk.push(root);
         while(!stk.isEmpty()){
             TreeNode cur = stk.pop();
             if(cur.left != null){
                 if(cur.left.left == null && cur.left.right == null){
                     res += cur.left.val;
                 }else{
                     stk.push(cur.left);
                 }
             }
             if(cur.right != null){
                 stk.push(cur.right);
             }
         }
         return res;
     }
 }

BFS也可以做.

Time Complexity: O(n). Space: O(n).

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public int sumOfLeftLeaves(TreeNode root) {
         if(root == null){
             return 0;
         }

         int res = 0;
         LinkedList<TreeNode> que = new LinkedList<TreeNode>();
         que.offer(root);
         while(!que.isEmpty()){
             TreeNode cur = que.poll();
             if(cur.left != null){
                 if(cur.left.left == null && cur.left.right == null){
                     res += cur.left.val;
                 }else{
                     que.offer(cur.left);
                 }
             }
             if(cur.right != null){
                 que.offer(cur.right);
             }
         }
         return res;
     }
 }