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;
    }
}