404. Sum of Left Leaves - jiejackyzhang/leetcode-note GitHub Wiki

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.

解题思路为preorder traversal，遇到左叶子就将其val加入res中。 可用recursion或iteration解。

##Approach 1: recursive

/**
 * 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);
            }
        }
        if(root.right != null) {
            res += sumOfLeftLeaves(root.right);
        }
        return res;
    }
}

##Approach 2: iterative

/**
 * 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> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if(node.left != null) {
                if(node.left.left == null && node.left.right == null) {
                    res += node.left.val;
                } else {
                    stack.push(node.left);
                }
            }
            if(node.right != null) {
                if(node.right.left != null || node.right.right != null) {
                    stack.push(node.right);
                }
            }
        }
        return res;
    }
}
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